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Introduction MASS and BALANCE

Oxford Aviation Academy (UK) Limited 2008All Rights ReservedThis text book is to be used only for the purpose of private study by individuals and may not be reproduced in any form or medium, copied, stored in a retrieval system, lent, hired, rented, transmited or adapted in whole or in part without the prior writen consent of Oxford Aviation Academy.Copyright in all documents and materials bound within these covers or atached hereto, excluding that material which is reproduced by the kind permission of third parties and acknowledged as such, belongs exclusively to Oxford Aviation Academy.Certain copyright material is reproduced with the permission of the International Civil Aviation Organisation, the United Kingdom Civil Aviation Authority and the European Aviation Safety Agency (EASA).This text book has been writen and published as a reference work to assist students enrolled on an approved EASA Air Transport Pilot Licence (ATPL) course to prepare themselves for the EASA ATPL theoretical knowledge examinations. Nothing in the content of this book is to be interpreted as constituting instruction or advice relating to practical fying.Whilst every efort has been made to ensure the accuracy of the information contained within this book, neither Oxford Aviation Academy nor the distributor gives any warranty as to its accuracy or otherwise. Students preparing for the EASA ATPL theoretical knowledge examinations should not regard this book as a substitute for the EASA ATPL theoretical knowledge training syllabus published in the current edition of CS-FCL 1 Flight Crew Licensing (Aeroplanes) (the Syllabus). The Syllabus constitutes the sole authoritative defnition of the subject mater to be studied in an EASA ATPL theoretical knowledge training programme. No student should prepare for, or is currently entitled to enter himself/herself for the EASA ATPL theoretical knowledge examinations without frst being enrolled in a training school which has been granted approval by an EASA authorised national aviation authority to deliver EASA ATPL training.Oxford Aviation Academy excludes all liability for any loss or damage incurred or sufered as a result of any reliance on all or part of this book except for any liability for death or personal injury resulting from Oxford Aviation Academys negligence or any other liability which may not legally be excluded.Cover Photograph: Airbus A380Photographed by Gabriel Savit, for www.airteamimages.com

Type of flight Baggage standard mass Domestic 11 kg Within the European 13 kg region Intercontinental 15 kg All other 13 kg (g) If an operator wishes to use standard mass values other than those contained in Tables 1 to 3 above, he must advise the Authority of his reasons and gain its approval in advance. He must also submit for approval a detailed weighing survey plan and apply the statistical analysis method given in Appendix 1 to JAROPS 1.620(g). After verification and approval by the Authority of the results of the weighing survey, the revised standard mass values are only applicable to that operator. The revised standard mass values can only be used in circumstances consistent with those under which the survey was conducted. Where revised standard masses exceed those in Tables 13, then such higher values must be used. (See IEM OPS 1.620(g).) (h) On any flight identified as carrying a significant number of passengers whose masses, including hand baggage, are expected to exceed the standard passenger mass, an operator must determine the actual mass of such passengers by weighing or by adding an adequate mass increment. (See IEM OPS 1.620(h) & (i).) (i) If standard mass values for checked baggage are used and a significant number of passengers check in baggage that is expected to exceed the standard baggage mass, an operator must determine the actual mass of such baggage by weighing or by adding an adequate mass increment. (See IEM OPS 1.620(h) & (i).) (j) An operator shall ensure that a commander is advised when a non-standard method has been used for determining the mass of the load and that this method is stated in the mass and balance documentation. JAROPS 1.625 Mass and balance documentation (See Appendix 1 to JAROPS 1.625) (a) An operator shall establish mass and balance documentation prior to each flight specifying the load and its distribution. The mass and balance documentation must enable the commander to [determine that the load and its distribution is such] that the mass and balance limits of the aeroplane are not exceeded. The person preparing the mass and balance documentation must be named on the document. The person supervising the loading of the aeroplane must confirm by signature that the load and its distribution are in accordance with the mass and balance documentation. This document must be acceptable to the commander, his acceptance being indicated by countersignature or equivalent. (See also JAROPS 1.1055(a)(12).) (b) An operator must specify procedures for Last Minute Changes to the load. (c) Subject to the approval of the Authority, an operator may use an alternative to the procedures required by paragraphs (a) and (b) above. INTENTIONALLY LEFT BLANKJAR-OPS 1.620(f)(2) (continued) JAR-OPS 1.625(a) (continued) 6JAR-OPS 1 SECTION 1 SUBPART J 1-J-4 01.03.98Appendix 1 to JAROPS 1.605 Mass and Balance General See JAROPS 1.605 (a) Determination of the dry operating mass of an aeroplane (1) Weighing of an aeroplane (i) New aeroplanes are normally weighed at the factory and are eligible to be placed into operation without reweighing if the mass and balance records have been adjusted for alterations or modifications to the aeroplane. Aeroplanes transferred from one JAA operator with an approved mass control programme to another JAA operator with an approved programme need not be weighed prior to use by the receiving operator unless more than 4 years have elapsed since the last weighing. (ii) The individual mass and centre of gravity (CG) position of each aeroplane shall be re-established periodically. The maximum interval between two weighings must be defined by the operator and must meet the requirements of JAROPS 1.605(b). In addition, the mass and the CG of each aeroplane shall be re-established either by: (A) Weighing; or (B) Calculation, if the operator is able to provide the necessary justification to prove the validity of the method of calculation chosen, whenever the cumulative changes to the dry operating mass exceed 05% of the maximum landing mass or the cumulative change in CG position exceeds 05% of the mean aerodynamic chord. (2) Fleet mass and CG position (i) For a fleet or group of aeroplanes of the same model and configuration, an average dry operating mass and CG position may be used as the fleet mass and CG position, provided that the dry operating masses and CG positions of the individual aeroplanes meet the tolerances specified in sub-paragraph (ii) below. Furthermore, the criteria specified in sub-paragraphs (iii), (iv) and (a)(3) below are applicable. (ii) Tolerances (A) If the dry operating mass of any aeroplane weighed, or the calculated dry operating mass of any aeroplane of a fleet, varies by more than 05% of the maximum structural landing mass from the established dry operating fleet mass or the CG position varies by more than 05 % of the mean aero-dynamic chord from the fleet CG, that aeroplane shall be omitted from that fleet. Separate fleets may be established, each with differing fleet mean masses. (B) In cases where the aeroplane mass is within the dry operating fleet mass tolerance but its CG position falls outside the permitted fleet tolerance, the aeroplane may still be operated under the applicable dry operating fleet mass but with an individual CG position. (C) If an individual aeroplane has, when compared with other aeroplanes of the fleet, a physical, accurately accountable difference (e.g. galley or seat con-figuration), that causes exceedance of the fleet tolerances, this aeroplane may be maintained in the fleet provided that appropriate corrections are applied to the mass and/or CG position for that aeroplane. (D) Aeroplanes for which no mean aerodynamic chord has been published must be operated with their individual mass and CG position values or must be subjected to a special study and approval. (iii) Use of fleet values (A) After the weighing of an aeroplane, or if any change occurs in the aeroplane equipment or configuration, the operator must verify that this aeroplane falls within the tolerances specified in sub-paragraph (2)(ii) above. (B) Aeroplanes which have not been weighed since the last fleet mass evaluation can still be kept in a fleet operated with fleet values, provided that the individual values are revised by computation and stay within the tolerances defined in sub-paragraph (2)(ii) above. If these individual values no longer fall Appendix 1 to JAR-OPS 1.605(a)(2) (continued) 7SECTION 1 SUBPART J JAR-OPS 1 01.03.98 1-J-5 within the permitted tolerances, the operator must either determine new fleet values fulfilling the conditions of sub-paragraphs (2)(i) and (2)(ii) above, or operate the aeroplanes not falling within the limits with their individual values. (C) To add an aeroplane to a fleet operated with fleet values, the operator must verify by weighing or computation that its actual values fall within the tolerances specified in sub-paragraph (2)(ii) above. (iv) To comply with sub-paragraph (2)(i) above, the fleet values must be updated at least at the end of each fleet mass evaluation. (3) Number of aeroplanes to be weighed to obtain fleet values (i) If n is the number of aeroplanes in the fleet using fleet values, the operator must at least weigh, in the period between two fleet mass evaluations, a certain number of aeroplanes defined in the Table below: Number of aeroplanes Minimum number of in the fleet weighings 2 or 3 n 4 to 9 n + 3 2 10 or more n + 51 10 (ii) In choosing the aeroplanes to be weighed, aeroplanes in the fleet which have not been weighed for the longest time shall be selected. (iii) The interval between 2 fleet mass evaluations must not exceed 48 months. (4) Weighing procedure (i) The weighing must be accomplished either by the manufacturer or by an approved maintenance organisation. (ii) Normal precautions must be taken consistent with good practices such as: (A) Checking for completeness of the aeroplane and equipment; (B) Determining that fluids are properly accounted for; (C) Ensuring that the aeroplane is clean; and (D) Ensuring that weighing is accomplished in an enclosed building. (iii) Any equipment used for weighing must be properly calibrated, zeroed, and used in accordance with the manufacturers instructions. Each scale must be calibrated either by the manufacturer, by a civil department of weights and measures or by an appropriately authorised organisation within 2 years or within a time period defined by the manufacturer of the weighing equipment, whichever is less. The equipment must enable the mass of the aeroplane to be established [accurately. (See AMC to Appendix 1 to JAR OPS 1.605 para(a)(4)(iii).)](b) Special standard masses for the traffic load. In addition to standard masses for passengers and checked baggage, an operator can submit for approval to the Authority standard masses for other load items. (c) Aeroplane loading (1) An operator must ensure that the loading of its aeroplanes is performed under the supervision of qualified personnel. (2) An operator must ensure that the loading of the freight is consistent with the data used for the calculation of the aeroplane mass and balance. (3) An operator must comply with additional structural limits such as the floor strength limitations, the maximum load per running metre, the maximum mass per cargo compartment, and/or the maximum seating limits. (d) Centre of gravity limits (1) Operational CG envelope. Unless seat allocation is applied and the effects of the number of passengers per seat row, of cargo in individual cargo compartments and of fuel in individual tanks is accounted for accurately in the balance calculation, operational margins must be applied to the certificated centre of gravity envelope. In determining the CG margins, possible deviations from the assumed load distribution must be considered. If free seating is applied, the operator must introduce procedures to ensure corrective action by flight or cabin crew if extreme longitudinal seat selection occurs. The CG Appendix 1 to JAR-OPS 1.605(a)(2) (continued) Appendix 1 to JAR-OPS 1.605(a)(4) (continued) 8JAR-OPS 1 SECTION 1 SUBPART J 1-J-6 01.03.98margins and associated operational procedures, including assumptions with regard to passenger seating, must be acceptable to the Authority. (See IEM to Appendix 1 to JAROPS 1.605 subparagraph (d).) (2) In-flight centre of gravity. Further to sub-paragraph (d)(1) above, the operator must show that the procedures fully account for the extreme variation in CG travel during flight caused by passenger/crew movement and fuel consumption/transfer. INTENTIONALLY LEFT BLANK INTENTIONALLY LEFT BLANK Appendix 1 to JAR-OPS 1.605(d)(1) (continued) 9Figure 1 - European region10JAR-OPS 1 SECTION 1 SUBPART J 1-J-8 01.03.98Appendix 1 to JAROPS 1.620(g) Procedure for establishing revised standard mass values for passengers and baggage (See IEM to Appendix 1 to JAROPS 1.620 (g)) (a) Passengers (1) Weight sampling method. The average mass of passengers and their hand baggage must be determined by weighing, taking random samples. The selection of random samples must by nature and extent be representative of the passenger volume, considering the type of operation, the frequency of flights on various routes, in/outbound flights, applicable season and seat capacity of the aeroplane. (2) Sample size. The survey plan must cover the weighing of at least the greatest of: (i) A number of passengers calculated from a pilot sample, using normal statistical procedures and based on a relative confidence range (accuracy) of 1% for all adult and 2% for separate male and female average masses (the statistical procedure, complemented with a worked example for determining the minimum required sample size and the average mass, is included in IEM OPS 1.620(g)); and (ii) For aeroplanes: (A) With a passenger seating capacity of 40 or more, a total of 2000 passengers; or (B) With a passenger seating capacity of less than 40, a total number of 50 x (the passenger seating capacity). [(3) Passenger masses. Passenger masses must include the mass of the passengers belongings which are carried when entering the aeroplane. When taking random samples of passenger masses, infants shall be weighed together with the accompanying adult. (See also JAR-OPS 1.620(c)(d) and (e).)](4) Weighing location. The location for the weighing of passengers shall be selected as close as possible to the aeroplane, at a point where a change in the passenger mass by disposing of or by acquiring more personal belongings is unlikely to occur before the passengers board the aeroplane. (5) Weighing machine. The weighing machine to be used for passenger weighing shall have a capacity of at least 150 kg. The mass shall be displayed at minimum graduations of 500 g. The weighing machine must be accurate to within 05% or 200 g whichever is the greater. (6) Recording of mass values. For each flight included in the survey, the mass of the passengers, the corresponding passenger category (i.e. male/female/children) and the flight number must be recorded. (b) Checked baggage. The statistical procedure for determining revised standard baggage mass values based on average baggage masses of the minimum required sample size is basically the same as for passengers and as specified in sub-paragraph (a)(1) (See also IEM OPS 1.620(g)). For baggage, the relative confidence range (accuracy) amounts to 1%. A minimum of 2000 pieces of checked baggage must be weighed. (c) Determination of revised standard mass values for passengers and checked baggage (1) To ensure that, in preference to the use of actual masses determined by weighing, the use of revised standard mass values for passengers and checked baggage does not adversely affect operational safety, a statistical analysis (See IEM OPS 1.620(g)) must be carried out. Such an analysis will generate average mass values for passengers and baggage as well as other data. (2) On aeroplanes with 20 or more passenger seats, these averages apply as revised standard male and female mass values. (3) On smaller aeroplanes, the following increments must be added to the average passenger mass to obtain the revised standard mass values: Number of passenger Required mass seats increment 1 5 incl. 16 kg 6 9 incl. 8 kg 10 19 incl. 4 kg Alternatively, all adult revised standard (average) mass values may be applied on aeroplanes with 30 or more passenger seats. Revised standard (average) checked baggage mass values are applicable to aeroplanes with 20 or more passenger seats. (4) Operators have the option to submit a detailed survey plan to the Authority for approval and subsequently a deviation from the revised standard mass value provided this deviating value is determined by use of the procedure explained 11SECTION 1 SUBPART J JAR-OPS 1 01.03.98 1-J-9 in this Appendix. Such deviations must be reviewed at intervals not exceeding 5 years. (See AMC to Appendix 1 to JAROPS 1.620(g), sub-paragraph (c)(4).) (5) All adult revised standard mass values must be based on a male/female ratio of 80/20 in respect of all flights except holiday charters which are 50/50. If an operator wishes to obtain approval for use of a different ratio on specific routes or flights then data must be submitted to the Authority showing that the alternative male/female ratio is conservative and covers at least 84% of the actual male/female ratios on a sample of at least 100 representative flights. (6) The average mass values found are rounded to the nearest whole number in kg. Checked baggage mass values are rounded to the nearest 0.5 kg figure, as appropriate. I NTENTIONALLY LEFT BLANK INTENTIONALLY LEFT BLANK 12JAR-OPS 1 SECTION 1 SUBPART J 1-J-10 01.03.98Appendix 1 to JAROPS 1.625 Mass and Balance Documentation See IEM to Appendix 1 to JAROPS 1.625 (a) Mass and balance documentation (1) Contents(i) The mass and balance documentation must contain the following information: (A) The aeroplane registration and type; (B) The flight identification number and date; (C) The identity of the Commander; (D) The identity of the person who prepared the document; (E) The dry operating mass and the corresponding CG of the aeroplane; (F) The mass of the fuel at take-off and the mass of trip fuel; (G) The mass of consumables other than fuel; (H) The components of the load including passengers, baggage, freight and ballast; (I) The Take-off Mass, Landing Mass and Zero Fuel Mass; (J) The load distribution; (K) The applicable aeroplane CG positions; and (L) The limiting mass and CG values. (ii) Subject to the approval of the Authority, an operator may omit some of this Data from the mass and balance documentation. (2) Last Minute Change. If any last minute change occurs after the completion of the mass and balance documentation, this must be brought to the attention of the commander and the last minute change must be entered on the mass and balance documentation. The maximum allowed change in the number of passengers or hold load acceptable as a last minute change must be specified in the Operations Manual. If this number is exceeded, new mass and balance documentation must be prepared. (b) Computerised systems. Where mass and balance documentation is generated by a computerised mass and balance system, the operator must verify the integrity of the output data. He must establish a system to check that amendments of his input data are incorporated properly in the system and that the system is operating correctly on a continuous basis by verifying the output data at intervals not exceeding 6 months. (c) Onboard mass and balance systems. An operator must obtain the approval of the Authority if he wishes to use an onboard mass and balance computer system as a primary source for despatch. (d) Datalink. When mass and balance documentation is sent to aeroplanes via datalink, a copy of the final mass and balance documentation as accepted by the commander must be available on the ground. INTENTIONALLY LEFT BLANK 13Chapter 1 JAR OPS 1 - ExtractQUESTIONS1. JAR legislation can be found in: a. JAR OPS-1 subpart A b. JAR OPS-1 subpart D c. JAR OPS-1 subpart K d. JAR OPS-1 subpart J2. The mass and centre of gravity of an aircraft must be established by actual weighing:a. by the pilot on entry of aircraft into service b. by the engineers before commencing servicec. by the operator prior to initial entry of aircraft into service d. by the owner operator before the frst fight of the day3. The operator must establish the mass of the Trafc Load:a. prior to initial entry into serviceb. by actual weighing or determine the mass of the trafc load in accordance with standard masses as specifed in JAR-OPS sub part J.c. prior to embarking on the aircraftd. by using an appropriate method of calculation as specifed in the JAR Ops subpart J4. The mass of the fuel load must be determined:a. by the operator using actual density or by density calculation specifed in the Operations Manual.b. by the owner using actual density or by density calculation specifed in JAR OPS - 1. c. by the pilot using actual density or by density calculation specifed in the Operations Manual.d. by the fuel bowser operator using actual density or by density calculation specifed in the Fuelling Manual.5. The Dry Operating Mass is the total mass of the aeroplane ready for a specifc type of operation and includes:a. Crew and passenger baggage, special equipment, water and chemicalsb Crew and their hold baggage, special equipment, water and contingency fuelc. Crew baggage, catering and other special equipment, potage water and lavatory chemicalsd. Crew and baggage, catering and passenger service equipment, potable water and lavatory chemicals.6. The Maximum Zero Fuel Mass is the maximum permissible mass of the aeroplane:a. with no useable fuelb. with no useable fuel unless the Aeroplane Flight Manual Limitations explicitly include it.c. including the fuel taken up for take-ofd. including all useable fuel unless the Aeroplane Flight Operations Manual explicitly excludes it.14Chapter 1 JAR OPS 1 - Extract7. The Maximum Structural Take-of Mass is:a. the maximum permissible total aeroplane mass on completion of the refuelling operation.b. the maximum permissible total aeroplane mass for take-of subject to the limiting conditions at the departure airfeld.c. the maximum permissible total aeroplane mass for take-of but excluding fuel. d. the maximum permissible total aeroplane mass at the start of the take-of run.8. The Regulated Take-of Mass:a. is the lower of maximum structural take-of mass and the performance limited take-of mass.b. is the higher of the maximum structural zero fuel mass and the performance limited take- of mass.c. the maximum structural take-of mass subject to any last minute mass changes.d. the maximum performance limited take-of mass subject to any last minute mass changes.9. The Take-of massa. the maximum permissible total aeroplane mass on completion of the refuelling operation.b. the mass of the aeroplane including everyone and everything contained within it at the start of the take-of run.c. the maximum permissible total aeroplane mass for take-of but excluding fuel. d. the maximum permissible total aeroplane mass at the start of the take-of run.10. The Operating Mass:a. is the lower of the structural mass and the performance limited mass b. is the higher of the structural mass and the performance limited mass c. is the actual mass of the aircraft on take-ofd. is the dry operating mass and the fuel load.11. The Basic Empty Mass is the mass of the aeroplane:a. plus non-standard items such as lubricating oil, fre extinguishers, emergency oxygen equipment etc.b. minus non-standard items such as lubricating oil, fre extinguishers, emergency oxygen equipment etc.c. plus standard items such as unusable fuids, fre extinguishers, emergency oxygen equipment, supplementary electronics etc.d. minus non-standard items such as unusable fuids, fre extinguishers, emergency oxygen and supplementary electronic equipment etc.15Chapter 1 JAR OPS 1 - Extract12. The Trafc Load:a. includes passenger masses and baggage masses but excludes any non-revenue load.b. includes passenger masses, baggage masses and cargo masses but excludes any non- revenue load.c. includes passenger masses, baggage masses, cargo masses and any non-revenue load. d. includes passenger masses, baggage masses and any non-revenue load but excludes cargo.13. The Operating Mass:a. is the take-of mass minus the trafc load.b. is the landing mass minus the trafc loadc. is the maximum zero fuel mass less the trafc loadd. is the take-of mass minus the basic empty mass and crew mass.14. The Trafc Load is:a. The Zero Fuel Mass minus the Dry operating Massb. The Take-of Mass minus the sum of the Dry Operating Mass and the total fuel load. c. The landing Mass minus the sum of the Dry Operating Mass and the mass of the remaining fuel. d. all the above15. The Basic Empty Mass is the:a. MZFM minus both trafc load and the fuel loadb. Take-of mass minus the trafc load and the fuel loadc. Operating mass minus the crew and fuel loadd. Landing mass less trafc load 16. Is it possible to fy a certifed aircraft at a Regulated Take-of mass with both a full trafc load and a full fuel load?a. It might be possible on some aircraft providing the mass and CG remain within limits b. Yes, all aircraft are able to do this.c. No, it is not possible on any aeroplane!d. Only if the performance limited take-of mass is less than the structural limited take-of mass.17. It is intended to fy a certifed aircraft loaded to the MZFM and MSTOM.a. The CG must be within limits during take-of and landing.b. The CG limits must be in limits throughout the fight, including loading/unloading.c. The CG does not have to be within limits during the whole of the fight. d. The CG does not have to be within limits during loading and unloading the aeroplane.16Chapter 1 JAR OPS 1 - Extract18. The term baggage means:a. Excess frieghtb. Any non-human, non-animal cargoc. any frieght or cargo not carried on the person d. personal belongings19. Certifed Transport category aircraft with less than 10 seats:a. may accept a verbal mass from or on behalf of each passenger.b. estimate the total mass of the passengers and add a pre-determined constant to account for hand baggage and clothing.c. may compute the actual mass of passengers and checked baggage.d. all the above.20. When computing the mass of passengers and baggage:1. Personal belongings and hand baggage must be included2. Infants must be classed as children it they occupy a seat3. Standard masses include infants being carried by an adult .4. Table 1, Table 2 and Table 3 must be used as appropriate if using standard masses for passengers and freight.5. Weighing must be carried out immediately prior to boarding and at an adjacent location.a. 1, 2 and v only b. 2 and 4 onlyc. 1, 2, 3 and 5 only d. All the above21. When computing the mass of passengers and baggage for an aircraft with 20 seats or more:1. Standard masses of male and female in Table 1 are applicable.2. If there are thirty seats or more, the All Adult mass values in Table 1 may be used as an alternative.3. Holiday Charter masses apply to Table 1 and Table 3 if the charter is solely intended as an element of a holiday travel package.4. Holiday fights and holiday charters atract the same mass values.a. 1, 3 and 4 only b. 1 and 2 onlyc. 3 and 4 onlyd. All the above17Chapter 1 JAR OPS 1 - Extract22. When computing the mass of passengers and baggage for an aircraft with 19 seats or less:1. The standard masses in Table 2 apply2. If hand baggage is accounted for separately, 6 kg may be deducted from the mass of each male and female.3. Table 2 masses vary with both the gender (male or female) of the seat occupant and the number of seats on the aircraft.4. Standard masses are not available for baggage. 5. Standard masses are not available for freight.a. 1 onlyb. 1, 2 and 4 only c. 3 and 5 onlyd. All the above23. When computing the mass of checked baggage for an aircraft with twenty seats or more:1. Table 1 applies 2. Table 2 applies 3. Table 3 applies4. Mass is categorised by destination. 5. Mass is categorised by gendera. 1, 3 and 4 only b. 2, 3 and 5 only c. 3 and 4 onlyd. All the above 24. On any fight identifed as carrying a signifcant number of passengers whose masses, including hand baggage, are expected to exceed the standard passenger mass the operator:a. must determine the actual mass of such passengersb. must add an adequate mass increment to each of such passengersc. must determine the actual masses of such passengers or add an adequate increment to each of such passengers.d. need only determine the actual masses or apply an increment if the Take-of mass is likely to be exceeded.25. If standard mass tables are being used for checked baggage and a number of passengers check in baggage that is expected to exceed the standard baggage mass, the operator:a. determine the actual masses of such baggageb. must determine the actual mass of such baggage by weighing or by deducting an adequate mass increment.c. need may no alterations if the Take-of mass is not likely to be exceeded.d. must determine the actual mass of such baggage by weighing or adding an adequate mass increment.18Chapter 1 JAR OPS 1 - Extract26. documentation:1. must be established prior to each fight2. must enable the commander to determine that the load and its distribution is such that the limits of the aircraft are not exceeded.3. must include the name of the person preparing the document.4. must be signed by the person supervising the loading to the efect that the load and its distribution is in accordance with the data on the document.5. must include the aircraft commanders signature to signify acceptance of the document.a. All the aboveb. 2, 4 and 5 only c. 1, 4 and 5 only d. 1 and 3 only27. Once the documentation has been signed prior to fight:a. no load alterations are allowed.b. documented last minute changes to the load may be incorporated.c. the documentation is not signed prior to fight.d. acceptable last minute changes to the load must be documented.28. Aircraft must be weighed:1. on initial entry into service2. if the records have not been adjusted for alterations or modifcations. 3. every four years after initial weigh4. whenever the cumulative changes to the dry operating mass exceed plus or minus 0.5% of the maximum landing mass.5. if the cumulative change in CG position exceeds 0.5% of the mean aerodynamic chord.a. 1 and 3 onlyb. 1, 2, 3, 4 and 5c. 1, 2 and 3 only d. 1, 3 and 5 only29. Aeroplane loading:1. must be performed under the supervision of qualifed personnel2. must be consistent with the data used for calculating the . 3. must comply with compartment dimension limitations4. must comply with the maximum load per running metre5. must comply with the maximum mass per cargo compartmenta. 1 and 2B onlyb. 1, 2, 4 and 5 only c. 1, 2, 3, 4 and 5d. 3, 4 and 5 only19Chapter 1 JAR OPS 1 - Extract30. An average dry operating mass and CG position may be used for a feet or group of aeroplanes:1. if they are of the same model and confguration2. providing the individual masses and CG positions meet specifc tolerances specifed in JAR OPS section J.3. providing the dry operating mass of any aeroplane does not vary by more than 0.5% of the maximum structural landing mass of the feet.4. providing that the CG position varies by more than 0.5% of the mean aerodynamic chord of the feet.5. providing appropriate corrections to mass and CG position are applied to aircraft within the feet which have a physical, accurately accountable diference.a. 1, 2, 3, 4 and 5b. 1, 2, 3 and 5 only c. 2, 3 and 4 onlyd. 1, 4 and 5 only20Chapter 1 JAR OPS 1 - ExtractANSWERS1 D 16 A2 C 17 B3 B 18 D4 A 19 D5 D 20 C6 B 21 B7 D 22 D8 A 23 C9 B 24 C10 D 25 D11 C 26 A12 C 27 D13 A 28 B14 D 29 C15 C 30 B21Chapter 2 Defnitions and CalculationsCHAPTER TWO DEFINITIONS AND CALCULATIONSContentsINTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23LIMITATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23EFFECTS OF OVERLOADING. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23EFFECTS OF OUT OF LIMIT CG POSITION. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24MOVEMENT OF CG IN FLIGHT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26SOME EFFECTS OF INCREASING AEROPLANE MASS . . . . . . . . . . . . . . . . . . . . . . . 27DEFINITIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27WEIGHING OF AIRCRAFT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32WEIGHING SCHEDULE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32MINIMUM EQUIPMENT LIST . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34CALCULATION OF FUEL MASS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34CALCULATION OF THE BASIC EMPTY MASS AND CG POSITION . . . . . . . . . . . . . . . . 37CALCULATION OF THE LOADED MASS AND CG POSITION FOR LIGHT AIRCRAFT . . . . 39CALCULATION OF THE LOADED MASS AND CG POSITION FOR LARGE AIRCRAFT . . . . 39COMPILING A DOCUMENT (LOAD SHEET) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39CG POSITION AS A PERCENTAGE OF MEAN AERODYNAMIC CHORD (MAC) . . . . . . . . 46RE-POSITIONING OF THE CENTRE OF GRAVITY . . . . . . . . . . . . . . . . . . . . . . . . . . 48RE-POSITIONING OF THE CENTRE OF GRAVITY BY REPOSITIONING MASS. . . . . . . . . . 48RE-POSITIONING OF THE CENTRE OF GRAVITY BY ADDING OR SUBTRACTING MASS . . 52GRAPHICAL PRESENTATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55CARGO HANDLING . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55FLOOR LOADING . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56LINEAR / RUNNING LOADS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57AREA LOAD LIMITATIONS. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57SINGLE ENGINE PISTON/ PROPELLER AIRCRAFT (SEP 1) . . . . . . . . . . . . . . . . . . . . . 59LIGHT TWIN PISTON/PROPELLER AIRCRAFT (MEP 1) . . . . . . . . . . . . . . . . . . . . . . . 59MEDIUM RANGE JET TWIN (MRJT 1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 CALCULATIONS (MRJT) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62LOAD AND TRIM SHEET MRJT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69Continued . . . . . .22Chapter 2 Defnitions and CalculationsContents ContinuedQUESTIONS FOR S.E.P.1; M.E.P.1 AND MRJT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75SELF ASSESSMENT QUESTIONS FOR MEP 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79SELF ASSESSMENT QUESTIONS FOR MRJT 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81ANSWERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9023Chapter 2 Defnitions and CalculationsINTRODUCTIONJAR-OPS 1 Subpart J requires that during any phase of operation the loading, mass and centre of gravity of the aeroplane complies with the limitations specifed in the approved Aeroplane Flight Manual, or the Operations Manual if more restrictive.It is the responsibility of the commander of the aircraft to satisfy himself that this requirement is met.LIMITATIONSLimitations on mass are set to ensure adequate margins of strength and performance, and limitations on CG position are set to ensure adequate stability and control of the aircraft in fight.EFFECTS OF OVERLOADINGThe four forces of lift, weight, thrust and drag acting on an aircraft all induce stress into the airframe structural members in the form of tension, compression, torsion, bending etc. The structure may, at the same time of absorbing these stress, be subject to extremes of temperature ranging from minus 56

c to plus 40

c. The stress and temperature factors gradually fatigue the structure as time progresses. Fatigue in this sense of the word, is a permanent loss of the physical properties (strength, durability, hardness etc) of the materials comprising the structure. Fatigue will, if left undetected or unatended, eventually cause the structure to fail altogether possibly with catastrophic and/or fatal consequences. Fatigue is cumulative and none reversible and the higher the fatigue level the greater the risk of premature structural failure. Structure that is inadvertently subject to additional fatigue may fail earlier than predicted or expected.The aircraft designer must, for each individual part of the structure, determine the frequency of application of the stress producing loads and, together with the temperature factors, determine the types of stress involved. Based on this data, a Design Limit Load (DLL) is calculated for each member and for the complete structure. The DLL is the maximum load that can be applied to the structure repeatedly during normal operations without inducing excessive fatigue and the pilot must never deliberately exceed this value. As a safe guard, the aviation authorities impose a factor of safety of 50% to the DLL to produce a Design Ultimate Load (DUL). The DUL is the minimum load the structure must be able to absorb in an emergency (heavier than normal landing or fight in exceptional gusty wind conditions) without collapsing. In order to keep weight to a minimum the aircrafts structure is manufactured from materials that are just capable of absorbing the DUL.Structure subject to loads in excess of the DUL is likely to sufer some permanent damage and may even collapse altogether.An aeroplanes principal function is to lift mass into the air, transport that mass through the air and then land it back on the ground without damage. Clearly, the greater the mass that has to be lifted the greater will be the loading on each member of the aircraft structure. Overloading the aeroplane will induce additional fatigue. For the purposes of cost efciency it is important to maximise the mass transported by the aeroplane but without overloading it.24Chapter 2 Defnitions and CalculationsThe manufacturer of the structural parts of the aircraft is responsible for determining the stresses the aeroplane will be subject to both on the ground and in the air, and to impose suitable mass limits so that the integrity of the structure is guaranteed throughout the aircrafts working life. The limits include: the maximum taxi mass (MTM); the maximum zero fuel mass (MZFM); the maximum structural take-of mass (MSTOM) and the maximum structural landing mass (MSLM). These values must never be exceeded in normal operation. It is necessary at this point to note that, in Mass & Balance terms, mass and weight are synonymous (used to express the same thing).Age, inappropriate handling, environmental and climatic conditions are all factors that induce stress and fatigue into the aircrafts structure. However, weight is the principal stress and fatigue inducing factor for aircraft. Weight also has pronounced efects on the aircrafts performance, handling and aerodynamic properties. With an increase in weight: Performance is reduced: Take of and landing distances will increase. V1 decision speed, VR rotation speed, V2 screen height, and the stopping distance will all increase The climb gradient, rate of climb and ceiling height will all reduce. The rate of descent will increase The stalling speed will increase and maximum speed will reduce. The safety margins and the efective speed range between low and high speed bufet will reduce. Drag and fuel consumption will increase Range and endurance will reduce. Wing root stresses will increase. Manoeuvrability will reduce. The aircraft will become less responsive to control inputs and more difcult to fy. Wing root stresses and undercarriage loads will increase as will tyre and brake wear.EFFECTS OF OUT OF LIMIT CG POSITIONThe centre of gravity (CG) is: the point that the total weight of the aircraft is said to act through the point of balance that part of the aircraft that follows the fight path the point that the aircraft manoeuvres about in the air the point that the three axes of the aircraft pass through.The position of the CG determines how stable or how manoeuvrable the aircraft will be. Starting at the mid position of the fuselage, a CG moving towards the nose of the aircraft will progressively increase the stability and, at the same time, progressively reduce the manoeuvrability. Similarly, a CG moving aft towards the tail of the aircraft will increase the manoeuvrability and decrease the stability. Too much stability increases the fying control stick forces and the work load on the pilot trying to overcome them. Too much manoeuvrability makes the aircraft unstable and difcult to control. 25Chapter 2 Defnitions and CalculationsWith regard to aeroplanes, the CG is not fxed. It moves in fight as a result of fuel burn, fap positions, and crew and passenger movements. It is the aircraft operators responsibility to ensure that the CG movement is retained with the limits imposed by the manufacturer.The manufacturer sets down CG range of movement limits to ensure that the average pilot is able to control the aircraft through all stages of fight without undue concern or fatigueThe following paragraphs indicate the efects that might occur if the CG is caused to exceed the limits. Students are advised to learn them well they are frequently asked in the exams.A CG outside the forward limit: Drag increases, consequently, fuel consumption, range and endurance decrease. In order to keep the nose of the aircraft from pitching downwards the tail plane must produce a balancing down load a bit like a see-saw. The resulting elevator defection increases drag, which in turn increase fuel consumption and reduces range and endurance. The longitudinal stability is increased, resulting in higher control column forces during manoeuvres and a corresponding increase in pilot fatigue. The increase in tail down force is equivalent to an increase in weight; consequently the stall speed will increase. An increase in stall speed has a signifcant efect on other performance aspects of the aircraft: take-of and landing speeds will increase, the available speed range will reduce and the safety margin between low and high speed bufet will narrow. The ability to pitch the nose up or down will decrease because of the increased stability. Take-of speeds V1, VR, VMU will increase. On the ground the aeroplane rotates about the main wheels and uses the elevators to raise the nose for take-of. The CG, being ahead of the main wheels, produces a down force that the elevators, together with the speed of the airfow passing over them, must overcome. The more forward the CG the greater the down force and, for a particular elevator defection, the greater the speed of the airfow required. The aircraft must accelerate for longer to produce the air speed required.A CG is outside the aft limit: Longitudinal stability is reduced and, if the CG is too far aft, the aircraft will become very unstable (like a bucking bronco). Stick forces in pitch will be light, leading to the possibility of over stressing the aircraft by applying excessive g. Recovering from a spin may be more difcult as a fat spin is more likely to develop. Range and endurance will probably decrease due to the extra drag caused by the extreme manoeuvres. Glide angle may be more difcult to sustain because of the tendency for the aircraft to pitch up.26Chapter 2 Defnitions and CalculationsMOVEMENT OF CG IN FLIGHTFigure 2.1 compares, in a simplistic arrow format, the efects on aircraft performance of having the CG on the forward CG limit to the performance that would be achieved with the CG on the aft limit. It goes on to shows how an increase in mass afects performance. Be advised that the M&B examinations contain a number of performance related theory type questions (they seldom include calculations).

= increase = decreaseThe JAR regulations state that the CG position must remain within the range limits at all times, whether in the air, taking of, landing or loading and unloading on the ground. Changes to the load distribution that may occur during any stage of the intended fight i.e. fuel burn, passenger or crew movements, will afect the CG position and must be properly accounted for prior to take-of.. DEFINITIONSThe following defnitions are to clarify and enhance the defnitions that are found in CAP 696, pages 3 and 4.Centre of GravityThe point through which the force of gravity is said to act on a mass (in CG terms, the point on the aircraft through which the total mass is said to act in a vertically downward manner). The Centre of Gravity is also the point of balance and as such it afects the stability of the aircraft both on the ground and in the air.Centre of Gravity limits. The CG is not a fxed point it has a range of movement between a maximum forward position and a maximum rearward position which are set by the aircraft manufacturer and cannot be exceeded. The CG must be on one of the limits or within the limit range at all times. The limits are given in the Flight manual and are defned relative to the datum. They may also be given as a percentage of the mean chord of the wing. The mean chord was known as the Standard mean Chord but is now known as the Mean Aerodynamic Chord or more simply, the MAC.DatumA point along the longitudinal axis (centre line) of the aeroplane (or it extension) designated by the manufacturer as the zero or reference point from which all balance arms begin. By taking moments about the balance arm the CG position of the aircraft can be determine. For the purposes of this phase of study the lateral displacement of the CG from the longitudinal axis is assumed to be zero.28Chapter 2 Defnitions and CalculationsBalance Arm The distance from the aircrafts Datum to the CG position or centroid of a body of mass. For example, the centroid of a square or rectangle is the exact centre of the square or rectangle and, in such cases, the balance arm is the distance from the datum to the exact centre of the square or rectangle. Unfortunately, cargo bays are seldom exact squares or rectangles and so the centroid (the point the total weight acts through) is given by the manufacturer.For the purposes of calculations, all balance arms ahead of (in front of) the datum are given a negative (-) prefx and those behind (aft of) the datum are given a positive (+) prefx.Figure 2.2 Positive and Negative Balance ArmsFigure 2.2 Positive and Negative Balance ArmsInexperienced students tend to get their positive and negative signs mixed up and thus fail to arrive at the correct answer. In arithmetical calculations, a positive value multiplied by a negative value results in a negative answer but two positive or two negative values multiplied together always produce a positive answer.Loading Index The result of multiplying a force by a mass often produces an answer of such magnitude that it is too bulky and time consuming to utilise. A Loading Index is simply a moment divided by a constant and has the efect of reducing the magnitude of the moment to one that is much easier to use. ITEM MASS (kg) ARM (in) MOMENT(kgin) CONSTANT INDEXBEM 31,994 691 22107854 /1000000 22FLT CREW 180 183 32940 /1000000 0CABIN CREW 540 1107 597780 /1000000 0.6SPECIAL EQUIPMENT12000 701 8412000 /1000000 8.4DOM = 44714 31150574 /1000000 31Figure 2.3 Example of an Index29Chapter 2 Defnitions and CalculationsIn the example shown, the moment for the DOM is 31150574 but the Dry Operating Index (DOI) is only 31. Later, as you progress through the course, you will be using a DOI and another index called the Fuel Index to complete a Load and trim Sheet.Basic Empty MassSee CAP 696 for the defnition of the BEM. All light aircraft use the BEM and its CG position as the foundation from which to calculate all relevant masses and CG positions see Figure 2.8 as an example. Dry Operating MassSee CAP 696 for the defnition of the DOM. All large aircraft use the DOM as the foundation from which to calculate all relevant masses and CG positions. The Load and Trim Sheet cannot be completed until the DOM and its CG position are known.Operating MassSee CAP 696 for the defnition of the OM. The OM is also used when completing the Load and Trim Sheet.Trafc LoadSee CAP 696 for the defnition of the Trafc Load. Originally known as the payload, the trafc load is the revenue generating load that pays the salaries and hopefully produces a proft for the operator. The defnitions of and calculations of the trafc load constitute a sizeable part of the exam. There are six ways to defne the trafc load and students need to be familiar with all of them.Useful LoadThe Useful Load is the sum of the trafc load and the take-of fuel load.The Maximum Zero Fuel MassSee CAP 696 for the defnition of the MZFM. The maximum stress in the wing roots occurs when the wing fuel tanks are empty. To ensure that the wings dont fold up permanently above the aircraft as the fuel is consumed a maximum zero fuel mass is imposed on the structure by the manufacturer.Maximum Structural Taxi MassEvery litre of fuel on board is essential for safe operations and aircraft are allowed to carry additional fuel for engine starting and ground taxiing purposes. This additional fuel, which is limited to a maximum value, is allowed to take the weight of the aircraft above the MSTOM during ground operations only. The additional fuel should be consumed by the time the aircraft is ready to commence the take-of run. The MSTM may also be referred to as the Ramp Mass or the Block Mass.See CAP 696 for defnitions of MSTOM, PLTOM, MSLM and PLLM30Chapter 2 Defnitions and Calculations

TOF

TL

ZFM

OM

UL

TOF

TL + + ++ = == = ==CREW & S/EQUIPT +

DOM

BEM ++ = RAMP OR BLOCK MASSSTART & TAXI FUEL

TOM =- TRIP FUEL

LM -=Note: Treat dotted line separately sNote: Treat doted lines separatleyFigure 2.4 Pieces of the Jigsaw31Chapter 2 Defnitions and CalculationsExamples of defnition questions:1. The operating mass of an aircraft is: a. The dry operating mass plus the take-of fuel mass b. The empty mass plus the take-of fuel mass c. The empty mass plus crew, crew baggage and catering d. The empty mass plus the trip fuel mass2. What efect has a centre of gravity close to the forward limit? a. A beter rate of climb capability b. A reduction in the specifc fuel consumption c. A reduce rate of climb for a particular fight path d. A decreased induced drag3. The DOM of an aeroplane is: a. TOM minus Operating Mass b. LM plus Trip Fuel c. Useful Load minus Operating Mass d. TOM minus Useful Load4. The Trafc Load of an aeroplane is: a. TOM minus Operating Mass b. LM plus Trip Fuel c. Useful Load minus Operating Mass d. TOM minus Useful Load.Answers are shown on Page 9032Chapter 2 Defnitions and CalculationsWEIGHING OF AIRCRAFTAircraft are weighed on specialised weighing equipment in a draught free hangar at periods specifed in JAR-OPS 1, Subpart J. On each and every occasion of weighing a WEIGHING SCHEDULE is compiled by the person in charge of the weighing procedure. The schedule lists the BASIC EQUIPMENT installed on the aeroplane and records the mass values displayed on the weighing apparatus, together with the calculated moments. It culminates in a statement defning the Basic Empty Mass and Centre of Gravity position of the aeroplane and is signed by the person in charge of the weighing procedure. The schedule is then retained in the aircrafts TECHNICAL LOG until the next weigh. At each subsequent weigh the list of basic equipment on the previous Weighing Schedule is used to defne the condition the aircraft must be reduced down to in order that an accurate comparison of weights can be determined. The weighing procedure is a time and manpower consuming process as all none basic items of equipment such as passenger seats and passenger service equipment (food and duty free trolleys etc) do not form part of the basic equipment and must frst be removed prior to the weigh and be refted afterwards. WEIGHING SCHEDULEA/C Type ______ Mark _____ Registration Number ______Registered owner _________ Date _____LIST OF BASIC EQUIPMENT ITEM MASS ARM MOMENT1. Pilots seat 100 1 1002. C/pilots seat 100 1 1003. Compass 25 2 504. S/B compass 25 2 505. Radio 20 2 40ITEM MASS (Kg) ARM MOMENT (kg in)Nose Wheel 134 -2 -268Left Main 550 90 49,516.5Right Main 550 90 49,516.5BEM 1,234 80 98,765Signed ___________________ Date _____________Figure 2.5. A Weighing Schedule (simplifed for training purposes)Any changes to the basic equipment that occur in the period between each weigh are recorded in the aeroplanes Technical Log and, as they vary from the equipment listed on the previous weighing schedule, they have to be accounted for separately at the next weigh.33Chapter 2 Defnitions and CalculationsWeighing Equipment. There are a number of ways of weighing aircraft accurately but, depending on the overall size and weight of the aeroplane, weigh-bridge scales, hydrostatic units or electronic equipment are the principal methods in use. Weigh-bridge scales. This equipment is generally used for light aeroplanes and consists of a separate electronic weighing platform for the nose or tail wheel and each main wheel assembly of the aircraft. The mass at each platform is recorded directly on the balance arm or electronic display and the masses are added together to give the BEM. Hydrostatic units. This equipment is used for larger, heavier aircraft and utilises the principle embodied in Pascals Law i.e. that the pressure of to a liquid in a closed container is proportional to the load applied. The units are fted at each jacking point and are interposed between the lifting jack and the jacking points on the aircraft. Again, the mass values on each unit are added together to give the BEM Electronic equipment. This equipment is also used on the larger, heavier aircraft and consists of strain gauges fted at each jacking point and utilises the principal that electrical resistance varies with the load applied. The readings are added together to give the BEM.The other masses and CG positions applicable to the aircraft e.g. DOM, OM, TOM etc, can be determined by simple addition and multiplication once the Basic Empty Mass and CG position of the aeroplane have been established. The mass of the fuel load can be calculated arithmetically providing the quantity and specifc gravity of the fuel are known.Actual mass of passengers and baggage can be used or the standard masses given in the tables in JAR Ops, Subpart J can be used as an expedient alternative.It is not normally possible to apply standard mass tables to freight because it varies so much from fight to fight. Therefore, in general, all freight has to be weighed. It is possible, in certain circumstances, for an operator to apply to the CAA for permission to produce and use standard mass tables for freight, but this exception is not part of the JAR examinations syllabus.Light aircraft use the BEM and CG position as the foundation from which the other mass and CG requirements are determined. Larger aircraft have additional mass and CG limits to comply with and they utilise a Load and Trim Sheet, incorporating the DOM as a basis, to simplify and standardise the process. The operator of feet aeroplanes of the same model and confguration may use an average DOM and CG position for the whole feet providing the requirements of JAR Ops, Subpart J are met. 34Chapter 2 Defnitions and CalculationsMINIMUM EQUIPMENT LISTAn aeroplane manufacturer must produce a Minimum Equipment List (MEL) for each of their aircraft type. The MEL defnes, amongst other things, the minimum level of serviceable usable equipment the aircraft must have prior to fight. As an example, a aeroplane with three engines would be permited to fy, in some circumstances, with only two engines operable. The MEL serviceable equipment requirements vary according to the climatic and environmental conditions that exist in various theatres of the world. An aircraft operator will extract from the MEL those limitations applicable to his/her aircraft and will enter them into the Aircraft Operating Schedule. In addition, the MEL lists the basic equipment requirements for each aircraft and also lists the optional specialist equipment that can be fted for a particular role. It is therefore very useful when determining the BEM and DOM of an aeroplane.CALCULATION OF FUEL MASSIt is the commander of the aeroplanes responsibility to ensure that there is sufcient fuel on board the aeroplane to safely complete the intended fight and to land with not less than a specifed level of fuel remaining in the tanks irrespective of delays and diversions.The safe operating fuel requirements defned above are satisfed by flling the tanks as shown in Figure 2.6.2% of the tank left empty for venting

2% of the tank left empty for venting TRIP FUEL Sufficient for flight from airfield a to airfield b together with enough extra fuel to allow for bad weather on route and/or landing delays at airfield b. ALTERNATE FUEL (Usually 3% to 5% of the trip fuel) Sufficient to allow for a diversion from airfield b to a planned diversion airfield c. FINAL RESERVE (Absolute emergency use only) CAPTAINS DISCRETION Economic or other reasons START AND TAXI FUEL Extended Twin Operations (ETOPs) for example Figure 2.6. Fuel Requirement Regulations35Chapter 2 Defnitions and CalculationsTaking of at airfeld a and landing at airfeld b is classed as a trip or sector. Having determined the mass of the fuel required for the trip, the commander of the aeroplane must convert this mass value into a quantity value for the beneft of the refuel operator. Fuel is usually dispenses in gallons or litres.In order to convert quantity (gallons or litres) into mass (lbs or kilograms) and visa versa, the density or the specifc gravity (sg) of the fuel must be known. Normally, the delivery note provided by the refuel operator provides the sg of the fuel taken up. However, if, for some unforeseen reason, the actual fuel density is not known a standard fuel density, as specifed by the operator in the Operations Manual, must be used. Density is defned as mass per unit volume and relative density or specifc gravity (sg), is simply a comparison between the mass of a certain volume of a substance and the mass an equal volume of pure water.The following chart is a handy method of converting volume to mass. Students are advised to remember the chart and how to use it as it will not be provided in the exams. Figure 2.3 Conversion ChartNB. Conversionfactors have beenrounded for simplicityhence small errorsmight occur.Figure 2.7 Quantity/Mass Conversion ChartWhen moving in the direction of the arrows multiply by the numbers above the line. When moving against the direction of the arrows divide by the numbers above the line.Note: Conversion factors have been rounded for simplicity hence small errors might occur.36Chapter 2 Defnitions and CalculationsWorked Example 1a) Find the mass of 50 Imperial gallons of AVGAS with a specifc gravity of 0.72. Mass = 50 x 10 x 0.72 = 360 lbb) For 50 US gallons this would be: Mass = 50 1.2 x 10 x 0.72 = 300 lbWorked Example 2Find the mass of 2250 litres of fuel with a density of 0.82.Mass = 2250 x 0.82 = 1845 kg.Try these on your own1. You require 63,000 kg of fuel for your fight, the aircraft currently has 12,000 kg indicated on the gauges. How many US gallons of fuel do you request if the density is 0.81.2. The refueller has metered 4596 Imperial Gallons; your fuel gauges indicated 5,600 lbs before refuelling. What should it indicate now? The fuel density is 0.79.3. If the mass of 6000 US gallons of fuel is 16780 kg, what is its S.G?4. The refuel bowser delivers 10,000 litres of fuel which is incorrectly entered on the aircraft load sheet as 10,000 kgs of fuel. Is the aircraft heavier or lighter than the take-of mass recorded on the Load Sheet and how would this efect the range? (Take the SG of the fuel as 0.75) a. Heavier and would decrease the range. b. Heavier and would increase the range c. Lighter and would decrease the range d. Lighter and would increase the rangeAnswers shown on Page 9037Chapter 2 Defnitions and CalculationsCALCULATION OF THE BASIC EMPTY MASS AND CG POSITIONIn order to determine the Basic Empty Mass and CG position of an aeroplane the aircraft must frst be prepared to the basic empty mass standard which entails removing all special equipment and useable fuel and oils.The aircraft is placed such that its main wheels and the nose (or tail) wheels rest on the individual weighing scales which have been calibrated and zeroed. The readings on the scales are recorded as shown:Figure 2.7 Basic Empty mass Weighing CalculationsFigure 2.8 Basic Empty Mass and CG CalculationsITEM MASS (lb) ARM (in) MOMENTNose wheel 500 -20 -10000L. Main wheel 2000 30 +60,000R. Main wheel 2000 30 +60,000BEM = 4500 lb Total Moment = +110,000 lb inCG = Total Moment = +110,000 lb in = +24.4 inches

Total Mass 4,500 inThe Basic Empty Mass of the aeroplane is 4500 lb and the CG is 24.4 inches behind the datum (as shown by the positive sign).The Basic Empty Mass is found by adding together the readings on the scales. To fnd the CG position we need to take moments about the datum. In Mass & Balance terms a moment is a mass multiplied by a balance arm. Remember that arms (distances) forward of the datum are negative and a negative multiplied by a positive gives a negative value (see the nose wheel line above).38Chapter 2 Defnitions and CalculationsNotice in the example that each of the three entries above the line consist of a mass multiplied by an arm to give a moment. The entry below the line consists of a mass and a moment but no balance arm. The missing arm is the CG position.To fnd the CG position the total moment is divided by the total mass.If the CG value is negative then the CG is in front of the datum otherwise it is behind the datum.It is important to distinguish between mass and weight. Mass is the amount of mater in a body in kilograms and weight is the force that the mater exerts on the earths surface, in Newtons.If the readings on the weighing scales are given in Newtons but the question asks for the BEM and CG position then it is necessary to convert the weight into mass to arrive at the right answer.ITEM WEIGHT (N) ARM (in) MOMENTNose wheel 500 -20 -10000L. Main wheel 2000 30 +60,000R. Main wheel 2000 30 +60,000 BEM = 4500 N Total Moment = +110,000 N inCG = Total Moment = + 110,000 N in = + 24.4 inches

Total Mass 4,500 inThe weight of the aeroplane is 4500 N but to fnd the Basic Empty Mass we must divide the weight by 9.81 m/s24500 N/ 9.81 m/s2 = 458.7 kgThe BEM = 458.7 kg and the CG is 24.4 inches behind the datumTry these examples yourself;1. An aeroplane with a two wheel nose gear and four main wheels rests on the ground with a single nose wheel load of 725 kg and a single main wheel load of 6000 kg. The distance between nose wheels and the main wheels is 10 metres. What is the BEM and how far is the centre of gravity in front of the main wheels? (25450 kg and 57 cm in front of the main wheels)2. A tail wheel aeroplane has readings of 2000 lbs and 2010 lbs for the main wheels and 510 lbs for the tail wheel. The nose wheel is 16 feet from the main wheels. What is the BEM and CG position? ( 1 foot = 12 inches). (4520 lb and 21.7 inches aft of the main wheels)3. A light aircraft has the datum 20 inches behind the nose wheel and 70 inches forward of the main wheels. The readings on the weighing scales are 255 N nose wheel and 1010 N on each main wheel. What is the BEM and CG position? (232 kg and 60 inches aft of the datum)39Chapter 2 Defnitions and CalculationsCALCULATION OF THE LOADED MASS AND CG POSITION FOR LIGHT AIRCRAFTThe loaded mass and CG is determined by tabulating the mass, arm and moment of the passengers, baggage, cargo, fuel and oil, adding the masses to the BEM to fnd the TOM and adding the moments to fnd the Total Moment. The CG is found by dividing the Total Moment by the Total Mass: CG = Total Moment/Total Mass.The LM is found by subtracting the fuel and oil consumed during the fight from the TOM. The CG position of the LM is found by taking moments and dividing the LM Moment by the LM. For simplicity and standardisation the mass, arms and moment data is tabulated on a Load Manifest or Load Sheet. (See Table 1, overleaf, for a complete example).CALCULATION OF THE LOADED MASS AND CG POSITION FOR LARGE AIRCRAFTThe TOM and ZFM and their respective CG positions are found by tabulating masses and moments as for light aircraft except the DOM is used as the starting point instead of the BEM.Because of the inclusion of start/taxi fuel and the complexity of the trafc load a Load and Trim sheet is used to formalise the mass and balance data.The ZFM and its CG position are used in preference to the LM and CG position because if the aircraft uses any of the contingency or alternate fuel, the actual landing mass and CG position may be signifcantly diferent to the calculated LM and CG values. If the TOM and ZFM are OK and their respective CGs are within limits then the CG will be in limits for any landing mass. (See pages 71 and 73 for examples of a completed Load Sheet and Trim Sheet respectively).COMPILING A DOCUMENT (LOAD SHEET)A load sheet in its simplest form is a list showing the BEM/DOM and CG position. Added in tabular form are the elements of the trafc load and fuel by their individual masses, arms and moments. From this list the take of mass and CG position can be calculated.A load sheet is individual to each type of aircraft and must be compiled before each fight.A load sheet is required by JAR-OPS 1 Subpart J to contain some mandatory information: The aeroplane registration and type The Flight Number The identity of the commander The identity of the person who prepared the document. The dry operating mass and CG position. The mass of take of fuel and trip fuel. The mass of consumables other than fuel. The items of trafc load, passengers, baggage, freight. The take of mass, landing mass and zero fuel mass. The load distribution. The aeroplane CG positions. The limiting mass and CG values.40Chapter 2 Defnitions and CalculationsIn the following pages we will concentrate on calculating the take of mass and CG position for the single engine piston aircraft SEP 1 using these values.Basic empty mass 2415 lbsFront seat occupants 340 lbs3rd and 4th seat passengers 340 lbsBaggage zone B 200 lbsFuel at engine start 60 US gallsTrip fuel (calculated fuel burn) 40 US gallsWhen completed the load sheet can be used to check that limiting values such as MZFM; RAMP MASS; MTOM and MLM have not been exceeded. The mass and CG limits are presented in graphical form (CAP 696, section 2, SEP 1 page 4, Figure 2.5) against which the calculated values can be checked.41Chapter 2 Defnitions and CalculationsITEM MASS (lbs) ARM (in) MOMENT/100Basic Empty Mass 2415 77.7 1876.46Front seat occupants 340 79 268.63rd and 4th seat pax 340 117 397.8Baggage zone A nil 1085th and 6th seat pax nil 152Baggage zone B 200 150 300Baggage zone C nil 180

SUB TOTAL = ZERO FUEL MASS 3295 2842.86Fuel loading 60 US galls 360 75 270SUB TOTAL = RAMP MASS 3655 3112.86Subtract fuel for start taxi -13 -10and run up. (see note) SUB TOTAL = TAKE OFF MASS 3642 3102.9Trip fuel -240 75 -180SUB TOTAL = LANDING MAS 3402 2922.9Table 1. Completed Load Sheet SolutionNote: Fuel for start, taxi and run up is normally 13 lbs at an average entry of 10 in the column headed moment / 100The arm data is entered on the load sheet in the appropriate columns as shown above. The individual moments are calculated by multiplying the mass of an item by its balance arm from the datum and entering the Figure in the moment column.Moments are often large numbers containing more than six digits and this can be an extra source of difculty. In the SEP example shown, the moments have been divided by one hundred to reduce the number of digits and make them more manageable. Take care if you use this procedure because you must remember at the end of any calculations to multiply the fnal answers by one hundred to arrive at the correct total moment.The fuel load may be given as a quantity (Imperial or American gallons) rather than a mass and you must convert it to mass before you can complete the load sheet (see para 9). Fuel mass and distribution may also be given in tabular form as shown in the example above, where the fuel mass and moment have been taken from the S.E.P.1 fuel chart, Figure 2.3, of the JAR Loading Manual.42Chapter 2 Defnitions and CalculationsThe Take-of and Landing CG positions can now be determined by using the following procedure:Take-of CG Sum (add up) the vertical MASS column to determine in turn, the ZFM, the Ramp Mass and the TOM. (ZFM = 3295 lb; Ramp Mass = 3655 lb and TOM = Ramp Mass Start/Taxi fuel = 3655 13 = 3642 lb) Check the Operating Manual to ensure that limiting masses of MZFM, MSTM and Regulated TOM have not been exceeded. Sum the vertical MOMENTS column to determine the total moment for ZFM, Ramp Mass moment and TOM. (ZFM moment = 284286 lb in; Ramp Mass moment = 284286 + 27000 = 311286 lb in; TOM moment 311286 1000 = 310286 lb in). Note, the Figure s in the table are shown in the abbreviated form i.e. 310286/100 = 3102.9 which is less accurate but easier to cope with. Divide the moment of the TOM by the TOM to determine the CG position at take-of. (TOM CG position = 310290 3642 = 85.2 inches aft of the datum). Check the Operating Manual to ensure that the CG is within limits at both the ZFM and TOM situations. If this is the case then the CG will remain within the limits throughout the fight and should not go out of limits during the journey provided the fuel is used in the correct sequence.Landing CG Determine the moment of the fuel used in fight (the trip fuel) by multiplying its mass by the fuel arm. In light aircraft the fuel arm will usually be the same as the one used previously to calculate the take-of CG position. However, caution is required because in some large aircraft the balance arm of the fuel may change with the quantity of fuel consumed. THE FUEL CONSUMED WILL GIVE A NEGATIVE MASS IN THE MASS COLUMN AND THIS WILL CHANGE THE MOMENT SIGN I.E. IF THE FUEL ARM IS POSITIVE THE FUEL MOMENT WILL BECOME NEGATIVE. In the MASS column, subtract the fuel used during the fight from the TOM to determine the Landing Mass. (Landing Mass = 3642 240 = 3402 lbs). Check the Operating Manual to ensure that the Regulated Landing Mass has not been exceeded. In the MOMENTS column, determine the sign of the fuel moment and add or subtract it as appropriate, to the TOM moment to determine the Landing moment. (Landing Moment = 310286 [240 x 75] = 310286 18000 = 292286 lb in [or 2922.9 lb in the abbreviated form]).43Chapter 2 Defnitions and Calculations If the CG is within limits at the ZFM then, for large aircraft, it is not normally necessary to calculate the Landing CG position because, as stated previously, the CG will be within limits throughout the fight. However, for light aircraft it is usual to determine the landing CG position and this is simply achieved by divide the Landing moment by the Landing Mass. (Landing CG position = 292286 3402 = 85.92 inches aft of the datum because it has a positive sign).The CG can also be found by using the Centre of Gravity Envelope for the SEP.1 shown below. This is a graphical representation of the mass and centre of gravity limits. The vertical axis is the mass in pounds, the horizontal axis is the CG position in inches aft of the datum and the slanted lines represent the moment/100.Fig 2.8 SEP 1 CG envelopeFigure 2.9 SEP 1 CG envelope44Chapter 2 Defnitions and CalculationsExample 3Try this example of calculating the take of and landing mass and CG position for a fctitious aircraft without using a formal load sheet.For the data given below: Find the CG for take of as loaded. Find the CG for landing after a fight lasting for 1 hour 30 minutes.Maximum Take-of Mass 2245 lbMaximum Landing Mass 2100 lbCentre of Gravity limits 2 in forward to 6 in aft of datumFuel Consumption 7.0 US Gallons per hourOil Consumption 1.0 US quart per hourITEM MASS (lb) ARM (in)Basic Mass 1275 -5Seats 1 and 2 340 -2Seats 3 and 4 170 30Fuel 35 US Gallons(SG 0.72)2Oil 8 US quarts(SG 0.9)-48Baggage 45 7045Chapter 2 Defnitions and CalculationsSolution:Lay out your solution in the same fashion as the load sheet;Calculate fuel and oil mass using the density specifed.Fuel: 35 1.2 x .72 x 10 = 210 lb (US galls to pounds)Oil : 8 4 1.2 x .9 x 10 = 15 lb (US qts to pounds)ITEM MASS (lb) ARM (in) MOMENTDOM 1275 -5 -6375Seats 1 and 2 340 -2 -680Seats 3 and 4 170 30 5100Fuel 35 US Galls 210 2 420Oil 8 US quarts 15 -48 -720Baggage 45 70 3150Take-of Mass 2055 lb Take-of Moment 895Take of CG = 0.435 inches aft of datum (within limits)To calculate landing CGFuel used in one and a half hours = 1.5 x 7 1.2 x .72 x 10 = 63 lbsOil used = 1.5 x .25 1.2 x .9 x 10 = 2.8 lbsLanding mass = 2055 - 63 - 2.8 = 1989.2 lbsLanding moment = Take of moment - Fuel used moment - Oil used momentLanding moment = +895 - (63 x +2) - (2.8 x -48)Landing moment = +895 - (126) - (-134) (minus and minus give plus) Landing moment = +895 - 126 + 134Landing moment = +903Landing CG = + 903

1989.2 = + 0.454 inches aft of datum (within limits)Note: Remember to check that the take-of mass and CG position and the landing mass and CG position are within the acceptable limits for the trip.46Chapter 2 Defnitions and CalculationsCG POSITION AS A PERCENTAGE OF MEAN AERODYNAMIC CHORD (MAC)In the previous examples the CG position and CG limits are given as distances from a datum. An alternative method is to state the CG position and its limits as a percentage of the Mean Aerodynamic Chord (MAC). This is common practice with many swept wing airliners and it is so with the twin jet we shall be studying next.The mean aerodynamic chord is one particular chord on the wing calculated from the aerodynamic characteristics of that particular wing. Because the CG afects many aerodynamic considerations, particularly stability, it is useful to know the CG position in relation to the aerodynamic forces.The length of the MAC is constant and it is at a fxed distance from the datum. The CG is located at some point along the MAC and the distance of the CG from the leding edge of the MAC is given in percentage forms - CG position of 25% MAC would mean that the CG was positioned at one quarter of the length of the MAC measured from the leading edge.The method of calculating the percentage of MAC is shown below.Figure 2 - 9Figure 2.10A = distance of CG from datum.B = distance of MAC leading edge from datum. C = length of MAC.The CG as a percentage of MAC = A - B x 100

C47Chapter 2 Defnitions and CalculationsExample 4If the MAC is 152 inc and its leading edge is 40 in aft of the datum, and the CG is 66 in aft of the datum, what is the CG position as a percent of MAC.

66 - 40 x 100 = 17.1% 152Now try these:-1. An aircraft has a MAC of 82 inches. The leading edge of the MAC is 103 inches aft of the datum. If the CG position is 14.7% MAC, what is the CG distance from the datum.2. If the CG position is 21% MAC, the MAC is 73 inches, and the CG datum is 26 inches aft of the leading edge of the MAC, what is the CG position relative to the datum.3. The CG limits are from 5 inches forward to 7 inches aft of the datum. If the MAC is 41 inches and its leading edge is 15 inches forward of the datum, what are the CG limits as % MAC.4. The MAC is 58 inches. The CG limits are from 26% to 43% MAC. If the CG is found to be at 45.5% MAC, how many inches is it out of limits.5. An aircraft of mass 62500 kg has the leading and trailing edges of the MAC at body stations +16 and +19.5 respectively (stations are measured in metres). What is the arm of the CG if the CG is at 30% MAC?48Chapter 2 Defnitions and CalculationsRE-POSITIONING OF THE CENTRE OF GRAVITYIf the centre of gravity is found to be out of limits for any part of the fight, the aircraft must not take of until the load has been adjusted so as to bring the centre of gravity into limits.This may be achieved in one of two ways: By re-positioning mass which is already on board the aircraft. This will usually be baggage or passengers. By adding or removing mass. Mass put on to the aircraft purely for the purpose of positioning or correcting the CG position is known as ballast.The minimum mass to be moved, or the minimum amount of mass to be loaded or of-loaded, will be that which just brings the centre of gravity on to the nearest limit. It may be preferable of course to bring the CG further inside the limits. When the amount of mass adjustment has been calculated, it must be ascertained that this makes the aircraft safe for both take-of and landing.RE-POSITIONING OF THE CENTRE OF GRAVITY BY REPOSITIONING MASS.Figure 2 -10 Repositioning MassFigure 2.11 Repositioning MassIn Figure 2.11 the centre of gravity has been found to be out of limits at a distance a inches aft of the datum. The forward CG limit is b inches aft of the datum. To bring the CG into limits, some baggage(m) will be moved from compartment A to compartment B.If a mass of m lb is moved from A to B the change of moment will be m x dThat is, the change of moment is equal to the mass moved (m) multiplied by the distance through which it moves (d).If the total mass of the aircraft is M, with the CG at a inches aft of the datum, the total moment around the datum is M x a.49Chapter 2 Defnitions and CalculationsIt is required to move the CG to (b) inches aft of the datum. The new total moment will then be M x b.The change in moment required is therefore M x b. - M x a = M (b - a) And M (b - a) = m x db - a is equal to the change to the CG position cc. and so m x d = M x ccThat is, the mass to move multiplied by the distance it moves is equal to the total aircraft mass multiplied by the distance the CG moves through.Example 5The CG limits of an aircraft are from -4 to +3 inches from the datum. It is loaded as shown below:ITEM MASS (lb) ARM (in) MOMENT (lb in)Basic Empty Mass 2800 2 5600Crew 340 -20 -6800Fuel 600 10 6000Forward Hold 0 -70 0Aft Hold 150 80 12000Total Mass = 3890 Total Moment = 16800Therefore CG = 16800 = 4.32 in

3890The Aft Limit is +3 and so the CG is 1.32 in out of limits (too far aft).It can be corrected by moving some freight or baggage from the rear hold to the forward hold, a distance of 150 ins. (80 ins aft to 70 ins forward).How much freight/baggage can be calculated by using our formula m x d = M x cc m x 150 = 3890 x 1.32 m = 3890 x 1.32

3290 = 1.69 in (in Limits)Now try these1. The CG limits of an aircraft are from 83 in. to 93 in. aft of the datum. The CG as loaded is found to be at 81 in. aft of the datum. The loaded mass is 3240 lb. How much mass must be moved from the forward hold, 25 in. aft of the datum, to the aft hold, 142 in. aft of the datum, to bring the CG onto the forward limit.2. An aircraft has a loaded mass of 5500 lb. The CG is 22 in. aft of the datum. A passenger, mass 150 lb, moves aft from row 1 to row 3 a distance of 70 in. What will be the new position of the CG. (All dimension aft of the datum)3. The loaded mass of an aircraft is 12 400 kg. The aft CG limit is 102 in. aft of the datum. If the CG as loaded is 104.5 in. aft of the datum, how many rows forward must two passengers move from the rear seat row (224 in. aft) to bring the CG on to the aft limit, if the seat pitch is 33 in. Assume a passenger mass of 75 kg each.4. An aircraft of mass 17400 kg, has its CG at station 122.2. The CG limits are 118 to 122. How much cargo must be moved from the rear hold at station 162 to the forward hold at station -100 (forward of the datum) to bring the CG to the mid position of its range?51Chapter 2 Defnitions and Calculations5. With reference to Figure 1 how much load should be transferred from No. 2 hold to No. 1 hold in order to move the CG from the out-of-limits value of 5.5 m to the forward limit value of 4.8 m. The total mass of the aircraft is 13 600 kg. Moving Mass Figure 1. Moving Mass Figure 16. With reference to Figure 2 the loaded mass of the aircraft is found to be 1850 lb and the CG moment 154 000 lb in. How much mass must be moved from the forward hold 40 in. aft of the datum, to the rear hold, 158 in. aft of the datum, to bring the CG on to the forward limit. Moving Mass Figure 252Chapter 2 Defnitions and CalculationsRE-POSITIONING OF THE CENTRE OF GRAVITY BY ADDING OR SUBTRACTING MASSThe position of the CG can also be adjusted by adding or subtracting mass. Mass added simply to re-position the CG is called ballast.Figure 2.11 Adding or Removing MassFigure 2.12 Adding or Removing MassTo calculate the minimum amount of ballast required :In Figure 2.12 the CG has been found to be out of limits at a distance X ins aft of the datum. The forward CG limit is at a distance Y ins aft of the datum. To bring the CG into limits, ballast will be put in compartment B, a distance Z ins aft of the datum.If the total mass of the aircraft is M lb., the total moment will be M x X lb in.If ballast of m lb is placed in compartment B in order to move the CG to its fwd limit, the total mass will increase to M + m and the new total moment will be (M + m) x Y . Assuming equilibrium to be maintained, the original total moment plus the moment of the added mass must equal the new total moment.Algebraically using the above notation then :- (M + m) x Y = (M x X) + (m x Z)New Total Moment = Old Total Moment + Cargo MomentThe same formula can be used for removing mass by changing the plus sign to a minus. So, for any calculation involving adding or subtracting mass remember the formula:-New Total Moment = Old total moment plus or minus the cargo momentNote that when calculating a change in CG position using the New Moment = Old Moment Change in Moment the distances (X, Y AND Z) are always measured from the datum itself.We will do an example calculation together.53Chapter 2 Defnitions and CalculationsExample 6The CG limits of an aircraft are from 84 ins to 96 ins aft of datum at all masses. It is loaded as shown below:ITEM MASS (lb) ARM (in) MOMENT (lb in)Basic Mass 1250 80 10000Crew 340 82 27880Fuel 300 72 22500Baggage 0 140 01890 150380 CG = 150380 = 79.6 ins aft of datum

56m = 148.5 Therefore mass of ballast required = 148.5 lbsIt would be necessary to check that loading the ballast did not cause the total mass to exceed the Maximum Take-Of Mass and as before, that the aircraft was in limits for landing.Although this may appear to be a long winded method it will always provide the correct answer, remember you may be asked to calculate a mass to add or remove, a change to the CG or the position to put ballast.54Chapter 2 Defnitions and CalculationsNow try these adding or removing mass problems:Question 1An aircraft has three holds situated 10 in 100 in and 250 in aft of the datum, identifed as holds A, B and C respectively. The total aircraft mass is 3500 kg and the CG is 70 in aft of the datum. The CG limits are from 40 in to 70 in aft of the datum. How much load must be removed from hold C to ensure that the CG is positioned on the forward limit.Question 2An aircraft has a mass of 5000 lb and the CG is located at 80 in aft of the datum. The aft CG limit is at 80.5 in aft of the datum. What is the maximum mass that can be loaded into a hold situated 150 in aft of the datum without exceeding the limit.Question 3The loaded mass of an aircraft is 108560 lb and the CG position is 86.3 ft aft of the datum. The aft CG limit is 85.6 ft. How much ballast must be placed in a hold which is located at 42 ft aft of the datum to bring the CG onto the aft limit.Question 4The aft CG limit of an aircraft is 80 in aft of the datum. The loaded CG is found to be at 80.5 in aft of the datum. The mass is 6400 lb. How much mass must be removed from a hold situated 150 in. aft of the datum to bring the CG onto the aft limit.Question 5An aircraft has a mass of 7900 kg and the CG is located at 81.2 in aft of the datum. If a package of mass 250 kg was loaded in a hold situated 32 in aft of the datum. What would the new CG position be?Question 6The CG limits of an aircraft are from 72 in. to 77 in. aft of the datum. If the mass is 3700 kg and the CG position is 76.5 in. aft of the datum, what will the change to the CG position be if 60 kgs is removed from the fwd hold located at 147 in fwd of the datum.Question 7An aeroplane has a zero fuel mass of 47800 kg and a performance limited take-of mass of 62600 kg. The distances of the leading edge and trailing edge of the MAC from the datum are 16m and 19.5m respectively. What quantity of fuel in Imperial gallons, must be taken up to move the CG from 30% MAC to 23% MAC if the tank arm is 16m aft of the datum and the fuel SG is 0.72?Answers shown on page 9055Chapter 2 Defnitions and CalculationsGRAPHICAL PRESENTATIONThe aircraft mass and CG position are frequently calculated using one of a number of graphical methods. These notes give several examples of the graphs that may be used: see CAP 696 and this book pages 43 and 73. There are two things in common on any of the graphs used: The CG must be within the envelope or on the line of the envelope. The mass of the aeroplane is always shown on the vertical scale.The example of a mass and CG envelope for the SEP 1, shown in CAP 696 Sect 2, SEP1 page 4, is unusual in that it uses both mass and moments on the vertical scale.The horizontal scale may use the CG position (inches, metres or centimetres), the moment of the Cg (kg inches, kg metres or kg centimetres) or the percentage of the CG along the mean aerodynamic chord. The MEP 1, envelope shown in CAP 696, Section 3 - MEP1 page 3, is an example of using CG position and the MRJT envelope shown in CAP 696, Section 4 - MRJT, page 9, is an example of the use of the MAC percentage.CARGO HANDLINGFigure 2.12 Cargo HandlingFigure 2.13 Cargo HandlingCargo compartmentsCompartments in the lower deck accommodate baggage and cargo. The compartments feature fre resistant sidewalls, ceilings and walkways. Cargo compartments are usually pressurised and heated and typically have fre detection and protection equipment. Cargo compartments usually have a maximum foor loading (kg/m2) and maximum running load value (kg/m).Containerised cargoBaggage and cargo can be loaded into standard size containers designed to ft and lock into the cargo compartment. The containers have an individual maximum mass limit and an individual foor loading limit (mass per unit area).56Chapter 2 Defnitions and CalculationsPalletised cargoCargo can also be loaded onto standard sized pallets and restrained with cargo nets or strops. Typically the forward area of the forward cargo compartment is conFigure d to take palletised freight.Bulk cargoBulk cargo can be loosely loaded in the area at the aft of the rear cargo compartment and separated from the containers by a restraining net atached to the foor ceiling and sidewalls.Cargo Handling Systems.The forward and aft cargo compartments typically have separate cargo power drive systems to move containers and cargo pallets.The power drive system is operated by a control panel at the door area of each cargo compartment and is capable of loading and unloading fully loaded containers or pallets in wet or dry conditions. A typical panel is shown below.Power drive units (PDU) are mounted in the foor and are controlled by the control joystick to move containers laterally and longitudinally in the doorway area and longitudinally in the compartment forward and aft of the doorway.Figure 2.13 Cargo bay Control PanelFigure 2.14 Cargo bay Control PanelGuides, rollers, stops, locks and tie down ftings are included in the appropriate places to provide adequate G restraint for the containers in fight.FLOOR LOADINGThe foors of the passenger cabin and the freight areas of the aircraft are limited in the load that they can carry. Placing excessive loads on the structure may not only cause visible panel creasing and local indentations but is likely to signifcantly accelerate structural fatigue. Fatigue is cumulative and can lead to major structural collapse of the structure with litle or no warning.The foor loading is defned both by linear loading and by area loading intensity57Chapter 2 Defnitions and CalculationsLINEAR / RUNNING LOADSThe linear (or running) loading limitation (lb per linear foot or kg per linear inch) protects the aircraft under foor frames from excessive loads. Depending on the units used, it is the total load permited in any one inch or one foot length of the aircraft (irrespective of load width). Figure 2.14 Linear Loading Figure 2.15 Linear LoadingCalculating the linear load distributionAs can be seen at Figure 2.14 (a), the linear load on the foor members is 500 kg divided by the 10 inch length i.e. 50 kg/inch; which is greater than the 45 kg/in allowed. However, the way in which an item of load is located in an aircraft can create an acceptable situation out of an unacceptable one.In example (b), by simply rotating the load through 90 the linear load on the foor members becomes 500 kg/20 inches i.e. 25 kg/in: which is well within the limit. It can be seen that the least linear loading occurs when the longest length is placed at right angles to the foor beams.AREA LOAD LIMITATIONSArea load limitations (kg/sq metre or lb/sq ft) protect the aircraft foor panels. The two permissible intensities of pressure are Uniformly Distributed (UD) and Concentrated loads.UD loadsThe general foor loading limitations for cargo are referred to as UD loads and are given as allowable lb/sq ft. Providing that a load (or series of loads) is within the allowable UD limitations for the foor area on which it rests, the loading is subject only to : not exceeding the linear load limitations. the individual and accumulated compartment load limitations.58Chapter 2 Defnitions and CalculationsConcentrated loadsLoad intensities which exceed the UD load intensities concentrated loads, can be carried providing approved load spreaders are used to distribute the load over sufcient area to ensure that loading limits are not exceeded.Load spreadersWhen concentrated loads or items of load with hard or sharp areas are carried on an aircraft, some form of foor protection is essential. The normal practice is to employ standard load spreaders of 2 inch thick timber.An example of load intensity and running load for you to do:An item of cargo has dimensions 3 ft x 4 ft x 12 ft and weighs 900 kg. Given that the maximum running load for the compartment is 20 kg/in and the maximum load intensity for the compartment is 70 kg/ft2 what are the running load values and the foor intensity values for the cargo and are there any limitations in the way in which it may be carried?(1 ft = 12 inches)Considering the RUNNING LOAD (underfoor protection) frst:Max running load = Load/ shortest length Mid running load = Load / mid length Min running load = Load / longest lengthMax running load = 900 kg = 25 kg/in ) Max = 20 kg/in

4 ft x 12 ft 48 ft2 ) Below limitThe cargo can be carried providing it is placed on the cargo-bay foor on either its medium area or largest area to prevent foor damage and that either the 4 feet length or the 12 feet length is placed parallel to the longitudinal axis of the aeroplane to prevent under-foor damage (spreading the load across the under foor frames).59Chapter 2 Defnitions and CalculationsSINGLE ENGINE PISTON/ PROPELLER AIRCRAFT (SEP 1)The procedure for compiling the documentation for the single engine piston/propeller aircraft (SEP1) given in CAP 696 is similar to the calculations previously specifed in these notes, but notice that the load manifest and CG envelope are specifc to that type of aircraft.A number of examples are included in the question section, starting on at page 75.LIGHT TWIN PISTON/PROPELLER AIRCRAFT (MEP 1)The procedure for compiling the documentation for the twin engine aircraft (MEP1) given in CAP 696 is also similar to the procedures defned in these note, but notice that the load manifest and CG envelope are specifc to that type of aircraft.A sample calculation based on compiling the load sheet for that aircraft is shown in CAP 696, MEP1, page 2. CAP 696, MEP page 1, contains specifc information based on the MAss & Balance requirements for that particular aircraft.These pages should be studied by the student after which the following self assessment questions should be atempted.A number of examples are included in the question section, starting on at page 75.MEDIUM RANGE JET TWIN (MRJT 1)The MRJT ( Medium Range Twin Jet) has a more complex loading and trim sheet which we shall deal with shortly.CAP 696, MRJT, pages 2 - 5, contain information specifc to the MRJT and should be studied in detail. Notes on particular items follow.MRJT Figure 4.1 and 4.2 Body StationThough it would be possible to locate components and parts of the airframe structure by actually measuring their distance from the CG datum, it would be difcult and impractical on anything other than a very small aircraft. Instead, aircraft are divided about their three axes by a system of station numbers, water lines and butock lines. The system of structural identifcation is not part of this subject except to say that in the past, station numbers were used as balance arms about the datum for frst-of-kind aircraft. However, new variants of original series aircraft are often made by inserting additional lengths of fuselage and consequently, the distances of components and structure from the original datum undergo a change. The station numbers could all be re- numbered to enable them to retain their use as balance arms but it is often more benefcial to retain the original station numbers as they are. This means that on some variant aircraft they can no longer be used as balance arms for CG purposes and in order to fnd out how far a particular station is from the datum a conversion chart is required.An example of such a chart is given in CAP 696, Section 4, MRJT, Page 1.60Chapter 2 Defnitions and CalculationsFigure 4.1 of the JAR Loading Manual shows a fuselage side view which includes the balance arms about the datum. The table at Figure 4.2 shows how the station numbers for the aircraft can be converted into balance arms and vice versa. An examination of the table will show that this aircraft is a variant of a previous series aircraft in that two fuselage sections 500A to 500G and 727A to 727G have been inserted. This is the reason the balance arms and the station numbers are not coincidental and why the conversion chart is required. Notice from the chart that the centre section of the airframe (stations 540 to 727) which is the same as the original aircraft, has retained its original station numbers and they are coincidental with the balance arms.Here are two examples of how to use the chart: Convert body station 500E into a balance arm. Body station 500E = 348 + 110 = Balance arm 458 inches. Convert balance arm 809 inches into a station number. Balance arm 809 82 = Station number 727Further examination of the chart shows that balance arm 809 is actually station number 727D.Now try these:1. What is the station number at the nose of the aircraft?2. What is the station number 1365 inches from the datum?3. What is the distance of station 500 from the datum?4. What is the distance of station 727C from the datum?MRJT Figure 4.3 and 4.4 Flap PositionThe movement of the faps on a large aircraft may have a considerable efect on the CG position. table 4.3 shows the moment change to the aircraft when the faps are extended or retracted,. for example retracting the faps from 30 to 0 would cause a total moment change of minus 15,000 kg.inches. Conversely extension of the faps from 0 to 40 would cause a total moment change of plus 16,000 kg. inches.The stabiliser seting for take of is extracted from the graph at Figure 4.4. The purpose of this is to allow the stabiliser trim to be set to allow the elevator sufcient authority to enable the aircraft to be rotated during the take of run and controlled during the frst stages of fight. The position of the CG will determine the stabiliser seting for take of.61Chapter 2 Defnitions and CalculationsFig 2.15 Stabiliser Trim SettingFigure 2.16 Stabiliser Trim SetingExample 1.What is the stabiliser trim seting if the CG is 15 % MAC and the faps are moved from the 5 to the 15 position?From the graph, if the CG is 15% MAC and the fap seting is 5 then the stabiliser trim seting will be 4.25 units nose up.If the CG was the same 15% MAC with a fap seting of 15 then the stabiliser trim seting would be 3.5 units nose up.Example 2.If, as a result of the trafc and fuel load, the CG moved from 15% MAC to 24% MAC what would be the change in stabiliser trim?From the graph, the stabiliser trim for 5 degrees take-of fap would change from 4.25 trim units at 15% MAC to 3 trim units at 24% MAC.Note: To enable the pilot to correctly set the stabiliser trim for take of there will be a trim indicator on the instrument panel or as an integrated part of an electronic display unit.62Chapter 2 Defnitions and CalculationsMRJT Figure 4.9 Cargo Compartment Limitation (CAP 696, MRJT, Page 5)These tables detail the cargo compartment limitations which must be considered when items of cargo are loaded to ensure that the limitations are not exceeded.Running LoadThe running load is the fore/aft linear load. For example a box having dimensions of 3 feet by 3 feet by 3 feet and weighing 100 kg would have a running load of 100 3 = 33.3 kg per foot which would be equal to 33.3 12 = 2.78 kg per inch. Be very careful to use the correct units(dont forget that there is a conversion chart on page 4 of the data sheets)However if a box having the same weight was 3ft x 2ft x 3ft then it could be positioned so that the 2ft side was running fore/aft in which case the running load would be:100 2 12 = 4.16 kg per inchThe cargo may have to be orientated correctly to prevent exceeding the running load limitations.Area LoadThe frst box would have a load intensity (mass per unit area) of 100 9 = 11.1 kg per square foot as the area it would be standing on would be 3 x 3 = 9 square feet.The second box however could be place on its side measuring 2ft x 3ft = 6 square feet area, it would therefore have a load intensity of 100 6 = 16.66 kg per sq.ftThe cargo may have to be orientated correctly to prevent exceeding the distribution load limitations. CALCULATIONS (MRJT)Calculating the Underload using the formulaeThe captain of the aircraft needs to know if he is able to accommodate any additional last minute changes to the load i.e. VIPs, emergency evacuation etc. Before the captain can allow any last minute additions to the load he must know if he has any spare load capacity or underload as it is referred to. Unfortunately, the underload is not simply the diference between the regulated take-of mass and the actual take-of mass, the MZFM and the Trafc Load have to be considered.Clearly, if the aircraft is already at its MZFM it has no underload. Even if it is below the MZFM, any underload may have already been consumed by extra fuel uptake.Any last minute load additions that are allowed will increase the size of the Trafc Load. The Trafc Load that can be carried is the lowest value of the Structural Trafc Load, the Take-of Trafc Load and the Landing Trafc Load.The allowed trafc load is the lowest of the following (you are required to know these formulas): Structural Limited Trafc Load = MZFM DOM Take-of Limited Trafc Load = RTOM DOM Take-of fuel Landing Limited Trafc Load = RLM DOM Fuel remaining63Chapter 2 Defnitions and CalculationsIn the case of the MRJT aircraft, if the DOM = 34300 kg, the take of fuel = 12000 kg and the fuel remaining at landing = 4000 kg:Structural Limited Trafc Load = 51300 34300 = 17000 kgTake-of Limited Trafc Load = 62800 34300 12000 = 16500 kgLanding Limited Trafc Load = 54900 34300 4000 = 16600 kgThe allowed Trafc load is thus 16,500 kg. But if the actual trafc load was only 16,000 kg there would be a 500 kg underload.You are required to know how to determine the allowable trafc load using the formulae above.Fuel Load DefnitionsStudents need to have a basic knowledge of the fuel load defnitions before they atempt trafc load calculations. Start and Taxi Fuel: The mass of fuel used in starting and operating the APU and the main engines and in taxying to the runway threshold for take of. It is assumed that at the point of releasing the brakes for take- of the aircraft is at or below the regulated take-of mass for the conditions prevailing. In operations where fuel is critical the start and taxi fuel must not be less than the amount expected to be consumed during the start and taxi procedures. Trip Fuel : This is the mass of fuel required to complete the take-of run, the climb, the cruise, the descent, the expected arrival procedures, the approach and landing at the designated airport. Contingency Fuel : Fuel carried in addition to the trip fuel for unforeseen eventualities such as avoiding bad weather or having an extended hold duration at the destination airport. the contingency fuel in calculations is usually given as a percentage of the trip fuel i.e. if the trip fuel is 1000 kg mass the contingency fuel at 5% of the trip fuel would be 50 kg. Dont forget that the contingency fuel is part of the landing mass if it is not actually used during the trip. Alternate (Diversion) Fuel : That mass of fuel required to carry out a missed approach at the destination airfeld, the subsequent climb out, transit to, expected arrival procedures, approach, descent and landing at an alternate airfeld. Final Reserve Fuel : The minimum fuel that should be in the tanks on landing. Essentially it is a fnal reserve for unplanned eventualities and should allow a piston engine aircraft to fy for a further 45 minutes or a jet engine aeroplane to fy for a further 30 minutes at a given height and holding speed. Additional Fuel : Only required if the sum of the trip, contingency, alternate and fnal reserve fuels are insufcient to cover the requirements of AMC OPS 1.255 (Instrument landings and power unit failures which not required for calculations).64Chapter 2 Defnitions and CalculationsThe take-of fuel for calculations is simply the sum of the above, excluding the Start and Taxi fuel.Note that the fuel state requirements vary with the intended fight plan and they are not always required.The landing fuel mass is the actual amount of fuel remaining in the tanks at touch down. In a trip where no eventualities have occurred it will include the contingency, alternate, fnal reserve and additional fuel masses if they were included in the fight plan.Calculating the Underload using the Load and Trim SheetThe Load and Trim Sheet used for the MRJT automatically calculates the underload but does it using a slightly diferent method than that shown above. Instead of determining the lowest trafc load the Load and Trim sheet frst determines the allowable Take-of Mass. It then calculates the allowable Trafc Load by deducting the Operating Mass from the allowable Take-of Mass. Finally, it calculates the underload by deducting the actual Trafc Load from the allowable Trafc Load.The allowable Take-of Mass is the lowest of: MZFM + Take-of Fuel Regulated Take-of Mass Regulated Landing Mass + Fuel used in fightFor example. In the case of the MRJT, the MZFM is given as 51300 kg, the MTOM is 62800 kg and the MLM is 54900 kg. Let us assume that there are no Performance Limits so that the Regulated Take-of and Landing masses are equal to the Structural Limited Take-of Mass and the Structural Limited Land Mass respectively. Let us also assume that the DOM is 34,000 kg, that the actual trafc load is 12,400kg, the take of fuel load is 16,000 kg and 8,000 kg of fuel was used in fight.Allowable Take-of mass is the lowest of: MZFM + Take-of Fuel 51,300 kg + 16,000 kg = 67,300 kg Regulated Take-of Mass 62,800 kg = 62,800 kg Regulated Landing Mass + Estimated fuel consumption 54,900 kg + 8,000 kg = 62,900 kgThe maximum allowable Trafc Load for the conditions prevailing can now be determined by subtracting the Operating Mass from the Regulated Take-of Mass i.e. :Maximum Trafc Load = Regulated Take-of Mass Operating Mass = 62,800 kg (34,000 kg + 16,000 kg) = 12,800 kg65Chapter 2 Defnitions and CalculationsThe underload can now be determined by subtracting the actual trafc load from the maximum allowable trafc load.Underload = maximum trafc load actual trafc load = 12,800 kg 12,400 kg = 400 kgFortunately, the MRJT Load and Trim Sheet makes easy work of the above calculations and you are advised to practice using the Load and Trim Sheet.Example calculations.Calculations may be carried out using the Loading Manifest ( data sheet Figure 4.10) and CG limits envelope (Figure 4.11) or the Load and Trim sheet (Figure 4.12).We will do a sample calculation using both methods.Using the following values complete the Loading Manifest Figure 4.10 and check the limiting values with the CG envelope.DOM 34,300 kg 15% MACPAX Total 116, standard weight 84 kg each 10 each in zone A and G 12 each in zone B and F 24 each in zone C, D and ECARGO 600 kg hold 1 1,500 kg hold 4 (includes checked baggage) FUEL 15,000kg at take of 260 kg start and taxi 10,000 kg trip fuelUse the CAP 696, MRJT, data sheets where required to fnd the balance arm for the MAC and vice-versa. The moment/1000 is calculated from the arm/1000. The balance arm is calculated by dividing the total moment by the total weight. The fuel balance arm and quantity in each tank is also found from the data sheets. Note: that the centre fuel tank content is used before the wing tank fuel content and that the centre tank includes 24 kg of unusable fuel.As a start, if the DOM CG position is 15% MAC, then that is 15% of 134.5 inches (CAP 696, MRJT, Page 2) or 20.175 inches. That makes the CG balance arm 20.175 + 625.6 = 645.8 inches aft of the datum The fuel load balance arm can be extracted from Figure 4.5 and 4.6 of the loading manual, for example the maximum contents of tanks one and two is 9084 kg with a balance arm of 650.7. Fill in this blank sheet using the above information then check the aircraft has not exceeded any of the limits in the CG envelopeSee if your answer agrees with mine! (I have estimated changes to the fuel tank CG position and accounted for the unusable fuel in the centre tank. However, the fuel CG position will be fxed and there will be no unusable fuel to account for in JAR exams)If you wish to check the mass values and CG positions using the load and trim sheet for theMRJT use a DOI of 40.5.66Chapter 2 Defnitions and CalculationsMaximum Permissible Aeroplane Mass ValuesTAXI MASS ZERO FUEL MASSTAKE-OFF MASS LANDING MASSITEM MASS (KG) B.A. InMOMENT KG-IN/1000CG % MAC1. DOM 34300 645.8 22150.9 15%2. PAX Zone A 284 -3. PAX Zone B 386 -4. PAX Zone C 505 -5. PAX Zone D 641 -6. PAX Zone E 777 -7. PAX Zone F 896 -8. PAX Zone G 998 -9. CARGO HOLD 1 367.9 -10. CARGO HOLD 4 884.5 -11. ADDITIONAL ITEMS -ZERO FUEL MASS12. FUEL TANKS 1&2 13. CENTRE TANKTAXI MASSLESS TAXI FUELTAKE OFF MASSLESS FLIGHT FUELEST. LANDING MASSFigure 2.17 LOADING MANIFEST - MRJT 167Chapter 2 Defnitions and CalculationsMaximum Permissible Aeroplane Mass ValuesTAXI MASS 63060kg ZERO FUEL MASS 51300kgTAKE-OFF MASS 62800kg LANDING MASS 54900kgITEMMASS (KG)B.A. InMOMENTKG-IN/1000CG %MAC1. DOM 34300 645.8 22150.9 15%2. PAX Zone A 840 284 238.5 -3. PAX Zone B 1008 386 389 -4. PAX Zone C 2016 505 1018 -5. PAX Zone D 2016 641 1292 -6. PAX Zone E 2016 777 1566 -7. PAX Zone F 1008 896 903 -8. PAX Zone G 840 998 838 -9. CARGO HOLD 1 600 367.9 221 -10. CARGO HOLD 4 1500 884.5 1327 -11. ADDITIONAL ITEMS -ZERO FUEL MASS 46144 649 29943.4 17.4%12. FUEL TANKS 1&2 9084 650.7 5911 -13. CENTRE TANK 6176 600.4 3708 -TAXI MASS 61404 644.3 39562.4 -LESS TAXI FUEL (C/TANK) -260 600.5 -156 -TAKE OFF MASS 61144 644.5 39406.4 14%LESS CENTRE TANK FUEL -5892 600.9 -3540.5 -LESS MAIN TANK FUEL -4108 628 -2580EST. LANDING MASS 51144 651 33286 18.9%Note: the fuel CG positions have been estimated for simplicity and allowance has been made for the 24 kg of unusable fuel in the centre tank. However, it is most unlikely that you will not be required to adjust for fuel tank CG changes or to account for unusable fuel quantities in JAR exams.Figure 2.18 LOADING MANIFEST - MRJT 1 (Table 3)68Chapter 2 Defnitions and Calculations

Figure 2.18. MRJT, Load and Trim SheetFigure 2.19 MRJT - Load and Trim Sheet69Chapter 2 Defnitions and CalculationsLOAD AND TRIM SHEET MRJTFigure 2.18 shows a combined Load and Trim sheet for a modern twin jet airliner designated JAA - Medium Range Jet Twin (MRJT). See CAP 696, MRJT1, Page 9, of the Loading Manual. (Students may be required to complete a part of the load and trim sheet during the JAA exams).The left hand side of the page (part A) is the loading document which itemises the mass and mass distribution within the aeroplane i.e. dry operating mass, trafc load and and fuel load. The right hand side (part B) indicates how each mass in turn subsequently afects the position of the C G in relation to the mean aerodynamic chord. The ZFM and the TOM must be within the relevant area of the graph envelop on completion of the mass calculations otherwise the aircraft is unsafe to fy.Part A (loading summary) should be completed as follows:-Section 1 is used to establish the limiting take of mass, maximum allowable trafc load and underload before last minute changes (LMC)Section 2 Shows the distribution of the trafc load using the following abbreviations. TR Transit B Baggage C Cargo M Mail Pax Passengers Pax F First Class Pax C Club/Business Pax Y EconomySection 3 Summarises the loading and is used to cross check that limiting values have not been exceeded.Part B is the distribution and trim portion. The lower part is the CG envelope graph. The vertical scale of which is given in terms of mass and the horizontal axis scale in terms of MAC. Using data from the loading summary enter the Dry Operating Index (DOI) for the DOM. Move the index vertically downwards into the centre of the frst row of horizontal boxes. Note that each box within the row has a pitch which represents either a defned mass or a number of persons. There is also an arrow indicating the direction in which the pitch is to be read. Move the index horizontally in the direction of the arrow to a pitch value corresponding to the value in the Mass/No box immediately to the left of the row. Repeat operations 1 to 3 above for each subsequent row of boxes down to and including row Og. After completing the index calculation for row Og drop a line vertically down until it bisects a mass on the vertical scale of the envelope corresponding to the Zero Fuel Mass. The point of bisection must occur within the MZFM envelope. Go to the horizontal row of boxes marked Fuel index and add the take-of fuel index.70Chapter 2 Defnitions and Calculations After completing the fuel index calculation drop a line vertically down from it until it bisects a mass on the vertical scale of the envelope corresponding to the Take-Of Fuel Mass. The point of bisection must occur within the TOM envelope. Providing the ZFM and the TOM are within the envelope as described above, the aircraft is safe for the intended fight - including any permited diversions.Example:The following example deals with part A and part B of the load sheet separately using the data shown. DOM = 34,300 kg DOI = 45.0 RTOM = 62,800 kg MZFM = 51,300 kg RLM = 54,900 kg Passengers 130 (Average mass 84 kg) Baggage 130 (@14 kg per piece) Cargo 630 kg Take of fuel 14,500kg Trip Fuel 8,500 kg Limitations for JAA - Medium Range Twin Jet Maximum Structural Taxi Mass 63,050 kg Maximum Structural Take Of Mass 62,800 kg Maximum Structural Landing Mass 54,900 kg Maximum Structural Zero Fuel Mass 51,300 kg Maximum Number of Passengers 141Cargo Hold 1 Max Volume 607 cu.ft Max Load 3,305 kg Hold 2 Max Volume 766 cu.ft Max Load 4,187 kgStandard Crew (Allowed for in DOM) Flight Deck 2 Cabin Crew forward 2 aft 1Section 1 of the load sheet is completed frst in order to fnd the three potential take of masses(a, b and c). The value at (a) is the take-of mass you would achieve if you loaded the aeroplane to the MZFM and then added your intended fuel load. The value at (b) is the Regulated take-of mass for the takeof airfeld conditions as existing and the value at (c) is the take-of mass you would achieve if you were to land at the Regulated landing mass and then added back the mass of the trip fuel. The lowest of (a), (b) and (c) is the limiting take-of mass for trafc load calculations. The maximum allowable trafc load for the trip can be determined by subtracting the operating mass from the limiting take-of mass. Subsequently, any underload can be calculated by subtracting the actual trafc load from the allowable trafc load (14,000 - 13,370 = 630). The underload sets the limiting mass for any last minute changes (LMC). Sections 2 and 3 of the load sheet detail the mass and distribution of the trafc load and give actual values of take-of and landing masses.71Chapter 2 Defnitions and CalculationsFig 2.19 Completed Trim SheetFigure 2.20 Completed Trim Sheet72Chapter 2 Defnitions and CalculationsIn part B the graph is entered at the top by drawing a vertical lie from the DOM index of 45 into the row for cargo compartment 1. This row is split into sections by heavy lines representing 1000kg , each section is split again into 10, each line representing 100 kg. the arrow in the box represents the direction to move to adjust for that mass.Follow the same procedure for each cargo compartment or seating compartment until have adjusted for all of the trafc load. Before adjusting for the fuel load draw the line down to intersect with the zero fuel mass to identify the ZFM CG position. Then adjust for the fuel index taken from the data sheets, page 30, and draw the line vertically down to identify the take-of CG position.Fig 2.20 Load and trim Calculation DiagramsFigure 2.21 Load and trim Calculation Diagrams73Chapter 2 Defnitions and CalculationsFigure 2.21Figure 2.2274Chapter 2 Defnitions and CalculationsExample.With regard to Part A above (Calculating the trafc/under load), a typical question is given below. For practice work out the answer by using both the load sheet method described on page 69 and the calculation method described on page 65.A scheduled fight of three hours estimated time, within Europe, is to be conducted. Using the data given calculate the maximum mass of freight that may be loaded in the following circumstances: Performance limited take-of mass 67,900 kg Performance limited landing mass 56,200 kg MZFM 51,300 kg DOM 34,960 kg Fuel on board at ramp 15,800 kg Taxi fuel 450 kg Trip fuel 10,200 kg Passengers (adults/each 84 kg) 115 (children/each 35 kg) 6 Flight crew (each 85 kg) 2 Cabin crew (each 75 kg) 5Allow standard baggage for each passenger (13 kg)a. 1,047 kgb. 6,147 kgc. 4,764 kgd. 4,647 kg75Chapter 2 Defnitions and CalculationsQUESTIONS FOR S.E.P.1; M.E.P.1 AND MRJTSELF ASSESSMENT QUESTIONS FOR SINGLE ENGINE PISTON/PROPELLER (SEP 1) Unless told otherwise, assume that the maximum fuel capacity is 74 gallons. For all questions refer to CAP 696 (Loading Manual) in Book 15.1. Where is the reference datum?a. 74 inches aft of the fwd CG position b. 80.4 inches aft of the rear CG position c. 87.7 inches aft of the rear CG position d. 39 inches forward of the frewall2. What are the CG limits?a. fwd limit = 74 inches to 80.4 inchesb. fwd limit = 74 inches, aft limit = 80.4 inches c. fwd limit = 74 inches, aft limit = 87.7 inchesd. fwd limit = 74 inches to 80.4 inches and aft limit = 87.7 inches3. What is the CG at the BEM?a. 77 inches b. 87 inchesc. 77.7 metres d. 77.7 inches4. What is the structural load limit for the foor at baggage zone C?a. 50 lb per square foot b. 100 lb per cubic foot c. 100 lb per square foot d. 100 kg per square inch5. What is the distance of the main undercarriage from the frewall?a. 97 inches b. 58 inchesc. 87.7 inches d. 39 inches6. The aircraft has six seats. Assuming no other cargo or baggage, what is the maximum fuel that can be carried if all six seats are occupied and the mass of each occupant is 180 lb?a. 50 lbs but the CG would be dangerously out of limits b. 155 lbs but the CG would be dangerously out of limits c. 50 lbs and the CG would be in limitsd. 155lbs and the CG would be in limits7. Where is the centroid of baggage zone B?a. 108 inches from the datum b. 120 inches from the datum c. 150 inches from the datum d. 180 inches from the datum76Chapter 2Defnitions and Calculations8. Assuming the weight and access is not a problem where can a box of mass 500 lb be positioned if the dimensions are 0.75 ft x 1.5 ft x 5 ft?a. in any of the baggage zones if placed on its smallest area b. in zones B or C if placed on its largest areac. in zone C only if placed on its middle area d. in zone A only if placed on its largest area9. Assuming the weight and access is not a problem, where can a cubic box of mass 500 lb be positioned if the dimensions are 3.15 ft?a. in any of the baggage zones b. in zone B or C onlyc. in zone A only d. in zone C only10. If the landing mass is 3155 lb and the trip fuel was 40 gallons, what was the ZFM if the fuel tanks held 60 gallons of fuel prior to take-of?a. 3001 lb b. 3035 lb c. 3098 lb d. 3111 lb11. What is the maximum ramp mass?a. 3650 lbs b. 3663 lbs c. 3780 lbs d. 3870 lbs12. How far is the main wheel from the aft CG limit?a. 0.7 inches behind the rear datumb. 0.7 inches forward of the rear datum c. 6.6 inches forward of the rear datum d. 9.3 inches aft of the rear datum13. How far is the frewall from the fuel tank centroid?a. 36 inches b. 37 inches c. 38 inches d. 39 inches14. If the total moment is less than the minimum moment allowed:a. useful load items must be shifted aftb. useful load items must be shifted forward c. forward load items must be increasedd. aft load items must be reduced77Chapter 2 Defnitions and Calculations15. The CG is on the lower of the fwd CG limits:a. at a mass of 2500 lb and moment of 185000 lb inb. at a moment of 175,000 lb in and a mass of 2350 lb c. at a moment of 192,000 lb in and a mass of 2600 lb d. all the above78Chapter 2Defnitions and Calculations79Chapter 2 Defnitions and CalculationsSELF ASSESSMENT QUESTIONS FOR MEP 11. What performance class does the aircraft belong to?a. Performance class A b. Performance class B c. Performance class C d. Performance class D2. Where is the reference datum?a. 78.4 inches forward of the wing leading edge at the inboard edge of the inboard fuel tankb. 25.3 inches forward of the nose wheelc. 109.8 inches forward of the main wheel d. all the above3. The main wheel is :a. 19 inches forward of the fwd CG limit at the maximum take-of mass b. 27.8 inches behind the fwd CG limit at a take-of mass of 3400 lbsc. 15.2 inches forward of the rear CG limit at the maximum take-of mass d. all the above4. The nose wheel is :a. 56.7 inches forward of the fwd CG limit at maximum take-of mass b. 65.5 inches forward of the fwd CG limit at maximum take-of mass c. 69.3 inches aft of the rear CG limit at maximum take-of massd. all the above5. What is the minimum fuel mass that must be consumed if the aircraft, having become airborne at maximum weight, decides to abort the fight.a. 1260 lb b. 280 lb c. 237 lbd. 202 lb6. If the pilot has a mass of 200 lb, what is the maximum trafc load?a. 1060 lbb. 1600 lbc. 1006 lbd. 6001 lb7. Assuming the maximum zero fuel mass and maximum take-of mass, what fuel load can be carried?a. 38.9 Imperial gallonsb. 46.6 US gallonsc. 176.8 litresd. any one of the above80Chapter 2Defnitions and Calculations8. A box of mass 100 lb is to be transported. The box dimensions are 9 x 9 x 12 inches. Which zones can it be carried in?a. all zones, both the mass and structural loading are within limitsb. zones 2 and 3 onlyc. no zones, both the mass and structural loading would be exceeded.d. no zones, the structural loading would be exceeded.9. A box of mass 360 lb is to be transported. The dimensions of the box are 1.7ft x 1.7ft x 1.8ft. Which zones can it be carried in?a. zones 2 and 3only but placed on the 1.7 x 1.7 faceb. zones 2 and 3 only but placed on the 1.7 x 1.8 facec. no zones, both the mass and structural loading would be exceededd. no zones, the structural loading would be exceeded.10. Assuming foor loading limits are acceptable, how much freight and fuel load can be carried for MTOM if the pilots mass was 200 lb?a. A full load in each zone plus 380 lb of fuelb. 50 lb in zones 1 or 4 but full loads in each of the other zones, plus 280 lbs of fuel.c. 350 lbs load in zone 4 but full loads in all the other zones, plus 280 lbs of fuel.d A full freight load in each zone plus 280 lb of fuel11. What is the maximum fuel tank capacity?a. not given.b. 123 US gallonsc. 46.6 US gallonsd. TOM minus ZFM12. If the aircraft is at MTOM with full fuel tanks and a pilot of mass 200 lb, what trafc load can be carried?a. nilb. 579 lbs providing at least 20.5 gallons of fuel are consumed in start, taxi and fight c. 625 lbs providing at least 43.3 gallons of fuel are consumed in start, taxi and fight d. 759 lbs providing at least 59.5 gallons of fuel are consumed in start, taxi and fight13. The CG when the TOM is 4300 lb and the corresponding moment is 408500 lb in is :a. 95 inchesb. 59 inchesc. 0.4 inches tail heavyd. 0.4 inches rear of the aft limit14. If the CG is 86 inches and the TOM is 4100 lb the aircraft is :a. just on the forward CG limitb. just outside the forward CG limitc. just inside the aft CG limitd. within the two forward limits81Chapter 2 Defnitions and CalculationsSELF ASSESSMENT QUESTIONS FOR MRJT 11. What is the total length of the fuselage?a. 1365 inchesb. 1375 inchesc. 1387 inchesd. 1395 inches2. How far is the front spar from the datum?a. 562 inchesb. 540 inchesc. 500 inchesd. 458 inches3. What is the distance between the two main access doors?a. 940 inchesb. 947 inchesc. 974 inchesd. 984 inches4. How far is the leading edge of the mean aerodynamic chord from the datum?a. 540 inches forward of the datumb. 589.5 inches forward of the datumc. 625.6 inches aft of the datumd. 627.5 inches aft of the datum5. What is the length of the mean aerodynamic chord?a. 104.5 inchesb. 114.5 inchesc. 124.5 inchesd. 134.5 inches6. What moment change occurs when the faps are fully retracted from the 15 degree position?a. a reduction of 14 kg inb. an increase of 14 kg inc. a reduction of 14000 kg ind. an increase of 14000 kg in7. What change in moment occurs when the faps are retracted from 40 degrees to 5 degrees?a. a negative moment of 5 kg inb. a negative moment of 11 kg inc. a negative moment of 16 kg ind. a negative moment of 5000 kg in82Chapter 2Defnitions and Calculations8. What stabiliser trim seting is required for take-of when the CG is 19% MAC for 5 degrees of take-of fap?a. 2.75b. 3.75c. 4.75d. 5.759. What is the maximum structural take-of mass?a. 63060 kgb. 62800 kgc. 54900 kgd. 51300 kg10. What is the CG range for maximum zero fuel mass?a. 8% MAC to 27% MACb. 12%MAC to 20% MACc. 7.5% MAC to 27.5% MACd. 8.5% MAC to 26% MAC11. Assuming the MZFM, what is the maximum allowable fuel mass for take-of?a. 10015 kgb. 10150 kgc. 11500 kgd. 15000 kg12. Assuming the standard masses have been used for both passengers and baggage, what is the mass of a full passenger and baggage load?a. 13027 kgb. 13677 kgc. 14127 kgd. 15127 kg13. What is the allowable hold baggage load for an aircraft with a full passenger complement?a. 1533 kg b. 1633 kg c. 1733 kg d. 1833 kg14. What is the underload if only maximum passenger hold baggage is carried?a. 3305 kg - 1833 kg = 1472 kgb. 4187 kg - 1833 kg = 2354 kgc. 7492 kg - 1833 kg = 5659 kgd. 9247 kg - 1833 kg = 7414 kg83Chapter 2 Defnitions and Calculations15. If the crew mass is 450 kg and the Zero Fuel Mass is 51300 kg, what is the Basic Empty Mass if a full trafc load is carried?a. 31514 kgb. 31773 kgc. 37713 kgd. 33177 kg16. Using the values for the data given in the Loading Manual, would the aircraft be able to carry both a full fuel load and a full Trafc load at take-of?a. Nob. Yes, providing the BEM was not more than 31145 kgc. Yes, providing the BEM was not less than 31451 kgd. Yes, providing the BEM was not more than 31514 kg17. If the DOM is given as 34300 kg and the aircraft has a full load of passengers and baggage, what additional cargo mass could it carry i.e. what is the underload?a. noneb. 3123 kgc. 3223 kgd. 3323 kg18. What is the maximum usable fuel quantity?a. 5311 US gallonsb. 5294 US gallonsc. 5123 US gallonsd. 5032 US gallons19. What is the maximum usable fuel mass?a. 16092 kg b. 16078 kg c. 16064 kg d. 16040 kg20. What is the allowable start and taxi fuel?a. 160 kgb. 260 kgc. 360 kgd. 460 kg21. What are the preferred zones for passenger loads if the pax load is low?a. zones E, F and Gb. zones C, D and Ec. zones B, C and Dd. A, B and C84Chapter 2Defnitions and Calculations22. How many seats are there in zone B?a. 15b. 18c. 21d. 2423. The leading edge of the MAC is given as 625.6 inches aft of the datum. What is the distance of the CG from the datum if it is found to be 16% of the MAC?a. 547 inchesb. 647 inchesc. 747 inchesd. 674 inches24. The CG is found to be 652.5 inches aft of the datum. What percentage is the CG to the MAC?a. 10%b. 15%c. 20%d. 25%25. If a passenger moves from a seat position corresponding to the balance arm at zone D to a position corresponding to the balance arm at zone F, what distance will the passenger have travelled and how many seat rows will he have passes?a. 255 inches and 8 seat rows b. 260 inches and 7 seat rows c. 265 inches and 6 seat rows d. 270 inches and 5 seat rows26. The balance arm for each of the seat zones is measured from the datum to:a. the front border line of the zone b. the centre line of the zonec. the rear border line of the zoned. the front border line of the next zone in sequence27. What is the maximum and minimum running load of a box of mass 500 kg and dimensions of 1m x 1.2m x 1.2m?a. 12.7 kg/in and 10.6 kg/in b. 10 kg/in and 12.4 kg/inc. 11 kg/in and 9.5 kg/ind. 15 kg/in and 13.1 kg/in28. What is the maximum and minimum distribution load intensity for a box of mass 500 kg and dimensions of 1m x 1.2m x 1.2m?a. 50.5 kg/sq ft and 40.6 kg/sq ftb. 47.3 kg/sq ft and 37.7 kg/sq ftc. 45.1 kg/sq ft and 35.8 kg/sq ftd. 38.7 kg/sq ft and 32.3 kg/sq ft85Chapter 2 Defnitions and Calculations29. All other parameters being acceptable, a box with a maximum and minimum running load of 12kg/in and 7 kg/in and a mass of 800 kg can be fted into:a. any compartment of either the forward or aft cargo compartmentb. the front section of the aft cargo compartment or the rear section of the forward cargo compartmentc. the rear section of the forward cargo compartment or the rear section of the aft cargo compartmentd. the centre section of forward cargo compartment only30. A box with a mass of 500 kg and dimensions 0.8m and 0.9m x 1.3m has a maximum and minimum distribution load intensity of:a. 64.6 kg/sq ft max and 39.7 kg/sq ft min b. 39.7 kg/sq ft max and 64.6 kg/sq ft min c. 44.7 kg/sq ft max and 39.7 kg/sq ft min d. 64.6 kg/sq ft max and 44.7 kg/sq ft min31. The maximum freight mass allowed is:a. 17017 lb b. 16517 lb c. 16017 lb d. 15517 lb32. Assuming all other parameters are acceptable, a box with a mass of 500 kg and with equal sides of 8.5 ft would ft into:a. either the front or rear cargo compartmentb. the forward cargo compartment onlyc. neither cargo compartmentd. the aft cargo compartment only33. The front compartment of the front cargo hold is situated below:a. passenger zone Ab. passenger zone Bc. passenger zone Cd. passenger zone D34. The balance arm of the centroid of the forward hold compartment is:a. 228 inchesb. midway between 228 inches and 286 inchesc. midway between 286 inches and 343 inchesd. 367.9 inches35. The maximum distribution load intensity for the cargo compartments is:a. 68 lb per sq ftb. 68 kg per sq metrec. 68 kg per sq ind. 68 kg per sq ft86Chapter 2Defnitions and Calculations36. Between 44,000 kg and 63,000 kg the rear CG limit as a percentage of the MAC:a. is constant at 28%b. increases from 28% to 29.5% c. decreases from 28% to 26% d. decreases from 28% to 9%Referring to the Load and Trim Sheet on page 29, answer questions 37 to 49 inclusive:37. The Trafc Load is:a. 39800 kg obtained from ZFM, 51300 kg less fuel mass 11500 kgb. obtained from the sum of pax mass plus baggage mass plus total cargo compartment massc. 13370 kg obtained from 10920 kg pax mass plus 2450 kg baggage mass plus 630 kg cargo massd. 13370 kg obtained from 10920 kg pax mass, 1820 baggage mass and 630 kg cargo mass38. The cargo distribution in section 4 is:a. 1220 kgb. 630 kgc. 1850 kgd. 1820 kg plus 630 kg39. The actual take-of mass is:a. 51300 kg ZFM plus 14500 kg take-of fuelb. 62800 kg less 8500 kg trip fuelc. 53670 kg less 14500 kg take-of fueld. 47670 kg ZFM plus 14500 kg take-of fuel mass40. The landing mass is:a. 62800 kg take-of mass less 8500 kg trip fuelb. 62170 kg take-of mass less 8500 kg trip fuelc. 62170 kg take-of mass plus 8500 kg trip fueld. 62800 kg take-of mass plus 8500 kg trip fuel41. In order to determine the underload the pilot starts by selecting the lowest mass from the three key masses given. The key masses are:a. Dry Operating Mass, Maxm Zero Fuel Mass and Take Of Massb. Maxm Zero Fuel Mass, Take Of Mass and Landing Massc. Dry Operating Mass, Maxm Zero Fuel Mass and Landing Massd. Trafc Load, Take-of Mass and Landing Mass42. From the Figure s given, if the actual take-of fuel mass (14500 kg) was added to the MaximumZero Fuel Mass the aircraft would be:a. Below the maximum Take-of mass by 350 kg b. Over the maximum Take-of mass by 530 kg c. Over the maximum Take-of mass by 3000 kg d. Below the maximum Take-of mass by 630 kg87Chapter 2 Defnitions and Calculations43. The actual underload for the aircraft after the trafc load and fuel load have been accounted for is:a. zerob. 720 kg c. 630 kg d. 960 kg44. What is the Dry Operating Index?a. 45 b. 12c. 54d. 1045. What are the seat row numbers in pax zone Oc?a. 4 6b. 6 8c. 7 10d. 8 -1346. What is the Take-of Mass as a percentage of the MAC?a. 18.3%b. 19.3%c. 20.3%d. 21.3%47. Prior to take-of there is a change in destination and so the pilot decides to take 2000 kg of fuel less. Using the Load and Trim Sheet, calculate the new Take-of mass and CG position.a. Cant be calculated because the landing mass will be too highb. 60800 kg Take-of mass and CG 17.5% MACc. 60170 kg Take-of mass and CG 18.8% MACd. 60170 kg Take-of mass and 19.3% MAC48. When adjusting the CG index for the fuel load, why is the line moved to the left as a minus index?a. Because the fuel will be consumed in fightb. Because the fuel is given a minus index in the fuel index correction tablec. Because the centroid of the tanks is behind the CG positiond. Because the graph would run out of range49. For a fuel mass of 11800 kg the index is:a. minus 4.5b. minus 5.7c. minus 6.3d. none of the above88Chapter 2Defnitions and Calculations50. A scheduled fight of three hours estimated fight time, within Europe, is being planned. Calculate the maximum mass of freight that may be loaded in the following circumstances: Structural limited take-of mass 62800 kg Structural limited landing mass 54900 kg MZFM 51300 kg Dry Operating Mass 34960 kg Fuel on board at ramp 15800 kg Taxi fuel 450 kg Trip fuel 10200 kg Passengers (adults each 84 kg) 115 Passengers (children each 35 kg) 6 Flight crew (each 85 kg) 2 Cabin crew (each 75 kg) 3 Standard baggage for each passenger 13 kga. 4647 kgb. 4102 kgc. 1047 kgd. 5545 kg89Chapter 2 Defnitions and Calculations90Chapter 2Defnitions and CalculationsANSWERSAnswers to defnition example questions1 a 2 c 3 d4 aAnswers to Fuel Mass Conversions1 = 16660 US gallons2 = 41908 lb3 = 0.744 = Lighter by 2500 kgAnswer to Basic empty mass and CG positionThe CG is 57 cm in front of the main wheels.Answers to Percentage Mean Aerodynamic Chord Problems1 = 115.054 inches2 = 10.67 inches fwd of datum3 = Fwd limit 24.3%, Aft limit 53.6%4 = 1.45 inches out of limits5 = 17.05mAnswers to Moving Mass Problems1 = 55.3816 lb2 = 23.9 inches3 = 7 rows4 = 146.1 kg5 = 952 kg6 = 43.2 lbAnswers to Adding or Removing Mass Problems1 = 500 kg2 = 35.97 lb3 = 1742.911 lb4 = 45.7 lb5 = 79.7 inches6 = 3.68 inches7 = 45.71 lb8 = 4455 Imperial gallonsAnswer to intensity and running load exampleThe boxes of cargo must go into hold B and the baggage into hold CAnswers to Station Numbers1 = Station 1302 = Station 12173 = 348 inches4. = 787 inches91Chapter 2 Defnitions and CalculationsAnswers to SEP1 Self Assessment Questions1 d 11 b2 d 12 d3 d 13 a4 c 14 a5 b 15 d6 b 16 b7 c8 b9 b10 bAnswer to example Trafc Load CalculationMaximum allowable freight mass = 4647 kgAnswers to MEP1. Self Assessment Questions1 b 6 a 11 b2 d 7 d 12 c3 b 8 d 13 d4 b 9 b 14 a5 c 10 aAnswers to MRJT Self Assessment Questions1 c 21 c 41 b2 b 22 b 42 c3 b 23 b 43 c4 c 24 c 44 a5 d 25 a 45 c6 c 26 b 46 a7 d 27 a 47 d8 b 28 d 48 b9 b 29 b 49 c10 a 30 a 50 c11 c 31 b12 b 32 a13 d 33 a14 c 34 d15 a 35 d16 a 36 c17 d 37 d18 b 38 b19 d 39 d20 b 40 b92Chapter 2Defnitions and Calculations93Chapter 3 Revision QuestionsCHAPTER THREEREVISION QUESTIONSContentsQUESTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95ANSWERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .104SPECIMEN EXAMINATION PAPER . . . . . . . . . . . . . . . . . . . . . . . . . . . .105ANSWERS TO SPECIMEN EXAMINATION PAPER . . . . . . . . . . . . . . . . . . .110DEBRIEF TO SPECIMEN EXAMINATION PAPER . . . . . . . . . . . . . . . . . . . .11194Revision Questions Chapter 395Chapter 3 Revision QuestionsQUESTIONS1. Defne the useful load:a. trafc load plus dry operating massb. trafc load plus usable fuel massc. dry operating mass plus usable fuel loadd. that part of the trafc load which generates revenue2. Determine the position of the CG as a percentage of the MAC given that the balance arm of the CG is 724 and the MAC balance arms are 517 to 1706.a. 14.2 %b. 15.3 %c. 16.3 %d. 17.4 %3. The distance from the datum to the CG is: a. the indexb. the momentc. the balance armd. the station4. Using CAP696, MRJT, fg 4.9. What is the balance arm, the maximum compartment load and the running load for the most aft compartment of the fwd cargo hold?a. 421.5 cm 3305 kg 13.12 kg per inchb. 1046.5 inches 711 kg 7.18 kg per kgc. 421.5 inches 2059 kg 13.12 kg per inchd. 1046.5 m 711 kg 7.18 kg per in5. If the maximum structural landing mass is exceeded: a. The aircraft will be unable to get airborneb. The undercarriage could collapse on landingc. No damage will occur providing the aircraft is within the regulated landing mass.d. No damage will occur providing the aircraft is within the performance limited landing mass.6. Use CAP 696, MRJT as appropriate. Prior to departure a MRJT is loaded with maximum fuel of 20100 ltr at an SG of 0.78. Calculate the maximum allowable trafc load that can be carried given the following data: PLTOM 6,200 kg PLLM 54,200 kg DOM 34,930 kg Taxi fuel 250 kg Trip fuel 9,250 kg Contingency and holding fuel 850 kg Alternate fuel 700 kg

a. 13,092 kgb. 12,442 kgc. 16,370 kgd. 16,842 kg96Revision Questions Chapter 37. Using CAP 696, fg 4.-12. Assuming the fuel index moves minus 5.7 from the ZFM index, what is the take-of CG as a percentage of the MAC?a. 20.1 %b. 19.1 %c. 23.0 %d. 18.2 % 8. For a conventional light aeroplane with a tricycle undercarriage confguration, the higher the take-of mass: 1. Stick forces at rotation will increase 2. Range will decrease but endurance will increase 3. Gliding range will reduce 4. Stalling speed will increase a. all statements are correctb. Statement 3 only is correctc. Statements 1 and 4 only are correctd. Statement 4 only is correct9. Due to a mistake in the load sheet the aeroplane is 1000 kg heavier than you believe it to be. As a consequence:a. V1 will be laterb. VMU will be later c. VR will be later d. V1, VMU, VR will all occur earlier 10. If the aeroplane was neutrally stable this would suggest that:a. the CG is forwardb. the CG is in mid rangec. the CG is on the rear limitd. the CG is behind the rear limit 11. The CG position is:a. set by the pilot b. set by the manufacturerc. able to exist within a ranged. fxed12. Which of the following would not afect the CG?a. Cabin crew members performing their normal duties.b. Fuel usagec. Stabilator trim setingd. Mass added or removed at the neutral point 97Chapter 3 Revision Questions13. Using the data for the MRJT in CAP 696, what is the CG as a percentage of the MAC if the CG is 650 inches from the datum.a. 17.03%b. 18.14%c. 19.25%d. 20.36%14. The CG datum has to be along the longitudinal axis:a. between the nose and the tail.b. between the leading and trailing edge of the MAC.c. but does not have to be between the nose and the tail.d. at the fre15. The CG is :a. the point on the aircraft where the datum is located.b. the point on the aircraft at which gravity appears to act.c. the point on the aircraft from where the dihedral angle is measured.d. the point on the aircraft where the lift acts through.16. The aircraft basic mass and CG position is found on :a. the weighing schedule and the aeroplane must be re-weighed if equipment change causes a change in mass or balanceb. On the loading manifest and is DOM trafc load.c. On the loading manifest and is ZFM useful loadd. On the weighing schedule and is adjusted to take account of any mass changes.17. When determining the mass of fuel/oil and the value of the SG is not known, the value to use is:a. determined by the operator (and laid down in the aeroplane OPS Manual. A pilot simply has to look it up)b. set out in JAR OPS Section 1c. determined by the aviation authorityd. determined by the pilot18. In mass and balance terms, what is an index?a. A cut down version of a force b. A moment divided by a constant c. A moment divided by a mass d. A mass divided by a moment19. Standard masses for baggage can be used when:a. 9 seats or moreb. 20 seats or morec. 30 seats or mored. less than 30 seats98Revision Questions Chapter 320. What is the zero fuel mass?a. MTOM minus fuel to destination minus fuel to alternative airfeld.b. Maximum allowable mass of the aircraft with no usable fuel on board.c. Operating mass minus the fuel load.d. Actual loaded mass of the aircraft with no usable fuel on board.21. If an aeroplane comes into lands below its MSLM but above the PLLM for the arrival airfeld: 1. Airframe structural damage will occur. 2. Tyre temperature limits could be exceeded. 3. It might not have sufcient runway length in which to stop safely. 4 A go-around might not be achievable. 5. Brake fade could occur..a. All the answers are correctb. 3 and 4 only are correctc. 2, 3, 4 and 5 only are correctd. 1, 3, 4 and 5 only are correct22. A twin engine aeroplane of mass 2500 kg is in balanced level fight. The CG limits are 82 in to 95 in from the nose position of the aeroplane and the CG is approximately mid range. A passenger of mass 85 kg, moves from the front seat 85.5 inches aft of the nose to the rear seat157.6 inches from the nose. What is the new CG position approximately?a. 2.5 inchesb. 87.5 inchesc. 91 inches d. 92.5 inches23. Calculate the Basic Empty mass and CG position for the MEP 1 shown below.

6 in 103.6 in 25.3 in 3450 N Left Main =5550 N Right Main = 5610N Datum a. BEM = 1489 kg and CG is 20 inches forward of datumb. BEM = 1456 kg and CG is 20 inches aft of the nosec. BEM = 1489 kg and CG is 20 inches aft of datumd. BEM = 1456 kg and CG is 89.6 inches aft of the nose6 in103.6 in 25.3 in3450 NLeft Main =5550 NRight Main = 5610NDatum99Chapter 3 Revision Questions24. A twin engine aeroplane is certifed for a MSTOM and a MLM of 58000 kg and 55000 kg respectfully. What is the limiting take-of mass for the aeroplane? PLTOM 61,000 kg PLLM 54,000 KG MZFM 3 6,000 kg Operating mass 55,000 kg Trip fuel 30,000 kg Contingency fuel 5% of trip fuel Alternative fuel 500 kg Final reserve 500 kg Flight duration 3 hours Fuel consumption 500 kg per hour per engine Useful load 41500 kga. 58,000 kg b. 61,000 kgc. 56,145 kgd. 56,545 kgRefer to CAP 696 for answers to 25, 26 and 2725. With reference to CAP 696 fgure 4.9, the centroid of the forward hold is :a. half way between stations 228 and station 500b. 314.5 inches forward of the aft cargo bay centroidc. 367.9 inches from the datumd. 367.9 inches from the nose of the aeroplane.26. The distance of the leading edge of the wing MAC from the datum is:a. undefnedb. 525.6 mc. 625.6 ind. 525.6 in 27. What is the CG as a percentage of the MAC of the fully loaded aircraft? BEM 12,000 kg Arm 3 m CG 25% MAC MAC 2 m Item Balance arm Front seats 2.5 m Rear seats 3 m Fuel @ 0.74 410 Ltr Fuel arm 2.5 m Rear seats Empty Pilot 80 kg Front seat Pax 80 kga. 16%b. 19%c. 21%d. 24%100Revision Questions Chapter 328. The maximum aircraft mass excluding all usable fuel is:a. fxed and listed in the aircrafts Operations Manual b. variable and is set by the payload for the trip.c. fxed by the physical size of the fuselage and cargo holds.d. variable and depends on the actual fuel load for the trip.29. Just prior to take-of, a baggage handler put an extra box of signifcant mass into the hold without recording it in the LMCs. What are the efects of this action? The aeroplane has a normal, tricycle undercarriage. 1. VMC will increase if the extra load is forward of the datum. 2. Stick forces in fight will decrease if the extra load is behind the datum. 3. Stick forces at VR will increase if the box is forward of the main wheels 4. VMU will occur later 5. The safe stopping distance will increase.a. 3, 4 and 5 onlyb. 2, 3 and 4 onlyc. 1 and 5 onlyd. all the above30. What is the maximum take-of mass given: MSTOM 43,000 kg MSLM 35,000 kg PLLM 33,000 kg MZFM 31,000 kg DOM 19,000 kg Total Fuel capacity 12,500 kg Maximum Trip Fuel 9,000 kg Contingency fuel 1000 kg Alternate fuel 500 kg Final reserve fuel 400 kga. 43,000 kgb. 42,000 kgc. 41,000 kgd. 40,000 kg31. What is the maximum mass an aeroplane can be loaded to before it moves under its own power?a. Maximum Structural Ramp mass b. Maximum Structural take-of massc. Maximum Regulated Ramp Massd. Maximum Regulated Take-of mass32. The weight of an aircraft in all fight conditions acts:a. parallel to the CGb. at right angles to the aeroplanes fight pathc. always through the MACd. vertically downwards101Chapter 3 Revision Questions33. With reference to MRJT Load and trim sheet (CAP696 Pg 31). If the DOM is 35000 kg and the CG is 14%, what is the D.O.I?a. 41.5b. 33c. 40d. 30 34. If the CG moves rearwards during fight:a. range will decreaseb. range will increasec. stability will increased. range will remain the same but stalling speed will decrease35. The CG of an aeroplane is situated at 115.8 arm and the mass is 4750 kg. A weight of 160 kg is moved from a hold situated at 80 arm to a hold at 120 arm. What would be the new CG arm?.a. 117.14b. 118.33c. 118.50d. 120.0136. What is the efect of moving the CG from the front to the rear limit at constant altitude, CAS and temperature?a. Reduced optimum cruise rangeb. Reduced cruise rangec. Increased cruise ranged. Increased stall speed.37. The baggage compartment foor-loading limit is 650 kg/m2. What is the maximum mass of baggage that can be placed in the baggage compartment on a pallet of dimensions 0.8m by 0.8m. The pallet has a mass of 6 kg?a. 416 kgb. 1,015 kgc. 650 kgd. 410 kg38. An aeroplane of 110,000kg has its CG at 22.6m aft of the datum. The CG limits are 18m to 22m aft of the datum. How much mass must be removed from a hold 30m aft of the datum to bring the CG to its mid point?a. 26,800 kgb. 28,600 kgc. 86,200 kgd. 62,800 kg39. Where does the mass act through when the aircraft is stationary on the ground?a. The centre of gravity b. The main wheelsc. It doesnt act through anywhere.d. The aerodynamic centre102Revision Questions Chapter 340. If an aircraft is weighed prior to entry into service who is responsible for doing the re-weigh to prepare the plane for operations?a. The manufacturer.b. The operator c. The pilotd. The fight engineer.41. An aeroplane has a tank capacity of 50000 Imperial gallons. It is loaded with fuel to a quantity of 165000 kg (790 kg/m3). What is the specifc gravity of the fuel and approximately how much more fuel could be taken up given that mass limits would not be exceeded?a. 0.73 46,053 gallonsb. 0.81 3,940 gallonsc. 0.72 46,000 gallonsd. 0.79 3,946 gallons42. Defne Balance Arm :a. BA = Mass / Momentb. BA = Moment / Massc. BA = Mass / Distanced. BA = Moment / Distance43. You have been given 16500 litres of fuel at SG 0.78 but writen down is 16500 kg. As a result you will experience :a. heavier stick forces at rotation and improved climb performance.b. heavier stick forces on rotation and distance to take-of increases.c. lighter stick forces on rotation and calculated V1 will be too high.d. lighter stick forces on rotation and V2 will be too low.103Chapter 3 Revision Questions104Revision Questions Chapter 3ANSWERS1 A 11 C 21 C 31 A 41 D2 D 12 C 22 C 32 D 42 C3 C 13 B 23 A 33 C 43 C4 C 14 C 24 A 34 B5 B 15 B 25 C 35 A6 B 16 A 26 C 36 C7 A 17 A 27 D 37 D8 C 18 B 28 A 38 B9 B 19 B 29 A 39 A10 D 20 D 30 B 40 B105Chapter 3 Revision QuestionsSPECIMEN EXAMINATION PAPER1. Defne the useful load:a. trafc load plus dry operating massb. trafc load plus usable fuel massc. dry operating mass plus usable fuel loadd. that part of the trafc load which generates revenue2. Determine the position of the CG as a percentage of the MAC given that the balance arm of the CG is 724 and the MAC balance arms are 517 to 1706.a. 14.2 %b. 15.3 %c. 16.3 %d. 17.4 %3. The distance from the datum to the CG is:a. the indexb. the momentc. the balance armd. the station4. Using CAP696, MRJT, fg 4.9. What is the balance arm, the maximum compartment load and the running load for the most aft compartment of the fwd cargo hold?a. 421.5 cm 3,305 kg 13.12 kg per inchb. 1046.5 inches 711 kg 7.18 kg per kgc. 421.5 inches 2,059 kg 13.12 kg per inchd. 1046.5 m 711 kg 7.18 kg per in5. Individual aircraft should be weighed in an air conditioned hangar :a. on entry into service and subsequently every 4 yearsb. when the efects of modifcations or repairs are not knownc. with the hangar doors closed and the air conditioning of.d. all the above.6. If a compartment takes a maximum load of 500 kg, with a running load limit of of 350 kg/m and a distribution load limit of 300 kg/m2 max. Which of the following boxes, each of 500 kg, can be carried? 1. 100 cm x 110 cm x 145 cm 2. 125 cm x 135 cm x 142 cm 3. 120 cm x 140 cm x 143 cm 4. 125 cm x 135 cm x 144 cma. any one of the boxes if loaded with due care as to its positioningb. either of boxes 2, 3 and 4 in any confgurationc. box 2 with its longest length perpendicular to the foor cross beam or box 3 in any confgurationd. either of boxes 3 and 4 with their longest length parallel to the aircraft longitudinal axis.106Revision Questions Chapter 37. Use CAP 696, Section 4, MRJT 1, as appropriate. Prior to departure a MRJT is loaded with maximum fuel of 20100 ltr at an SG of 0.78. Calculate the maximum allowable trafc load that can be carried given the following data:PLTOM 67,200 kgPLLM 54,200 kgDOM 34,930 kgTaxi fuel 250 kgTrip fuel 9,250 kgContingency and holding fuel 850 kgAlternate fuel 700 kga. 13,092 kgb. 1,2442 kgc. 16,370 kgd. 16,842 kg8. If the maximum structural landing mass is exceeded:a. The aircraft will be unable to get airborneb. The undercarriage could collapse on landingc. No damage will occur providing the aircraft is within the regulated landing mass.d. No damage will occur providing the aircraft is within the performance limited landing mass.9. For a conventional light aeroplane with a tricycle undercarriage confguration, the higher the take-of mass (assume a stab trim system is not fted): 1. Stick forces at rotation will increase 2. Range will decrease but endurance will increase 3. Gliding range will reduce 4. Stalling speed will increasea. all statements are correctb. Statement 3 only is correctc. Statements 1 and 4 only are correctd. Statement 4 only is correct10. Due to a mistake in the load sheet the aeroplane is 1000 kg heavier than you believe it to be. As a consequence:a. V1 will be laterb. VMU will be laterc. VR will be laterd. V1, VMU, VR will all occur earlier11. If the aeroplane was neutrally stable this would suggest that:a. the CG is forwardb. the CG is in mid rangec. the CG is on the rear limitd. the CG is behind the rear limit107Chapter 3 Revision Questions12. The CG position is:a. set by the pilotb. set by the manufacturerc. able to exist within a ranged. fxed13. Which of the following would not afect the CG position?a. Cabin crew members performing their normal duties.b. Fuel consumption during fight.c. Horizontal stabilator trim setingd. Mass added or removed at the neutral point14. An aircraft is about to depart on an oceanic sector from a high elevation airfeld with an exceptionally long runway in the tropics at 1400 local time. The regulated take-of mass is likely to be limited by :a. MZFMb. Obstacle clearancec. Maximum certifed Take-of massd. Climb gradient15. An aircraft is fying at 1.3 VS1g in order to provide an adequate margin above the low speed bufet and transonic speeds. If the1.3VS1g speed is 180 kts CAS and the mass increases from285000 kg to 320000 kg, What is the new 1g stalling speed?a. 146.7 kts, drag will increase and nautical mile per kg fuel burn will decrease.b. 191 kts, drag will increase and range nm/kg will increase.c. 191 kts, drag will increase and nm/kg fuel burn will decrease.d. 147 kts, drag will remain the same and nm/kg fuel burn will increase16. The datum for the balance arms has to be along the longitudinal axis:a. between the nose and the tail.b. between the leading and trailing edge of the MAC.c. but does not have to be between the nose and the tail.d. at the fre wall.17. The useful load is:a. TOM fuel massb. BEM plus fuel loadc. TOM minus the DOMd. TOM minus the operating mass18. In Mass & Balance terms, what is an index?a. A cut down version of a forceb. A moment divided by a constantc. A moment divided by a massd. A mass divided by a moment108Revision Questions Chapter 319. Standard masses for baggage can be used for aircraft with:a. 9 seats or moreb. 20 seats or morec. 30 seats or mored. less than 30 seats20. If an aeroplane comes into lands below its MSLM but above the PLLM for the arrival airfeld: 1. Airframe structural damage will occur. 2. Tyre temperature limits could be exceeded. 3. The runway length might be inadequate. 4 A go-around might not be achievable. 5. Brake fade could occur.a. 1 and 5 onlyb. 3 and 4 onlyc. 2, 3, 4 and 5 onlyd. 1, 3, 4 and 5 only21. What is the zero fuel mass?a. MTOM minus fuel to destination minus fuel to alternative airfeld.b. Maximum allowable mass of the aircraft with no usable fuel on board.c. Operating mass minus the fuel load.d. Actual loaded mass of the aircraft with no usable fuel on board.22. An aeroplane develops a serious maintenance problem shortly after take-of and has to return to its departure airfeld. In order to land safely the aircraft must jetison fuel. How much fuel must be jetisoned.a. Sufcient to reduce the mass to the zero fuel massb. The pilot calculates the amount of fuel to jetison to reduce the mass to a safe level at or below the RLM.c. The fuel system automatically stops the jetison at the RLM.d. As much as the pilot feels is just insufcient to land safely23. Calculate the amount of cargo that could be loaded into the aircraft given the following information and using the CAP 696, Section 4, MRJT 1, as necessary.Dry Operating Mass 34,900 kgPerformance Limited Landing Mass 55,000 kgTrip Fuel 9,700 kgContingency Fuel 1,200 kgAlternate Fue 1,400 kg130 passengers at 84kg each 10,920 kg130 bags at 14kg each 1,820 kga. 2,860 kgb. 3,660 kgc. 4,660 kgd. 5,423 kg

1. MZFM DOM = 51,300 34,930 = 16,370 kg 2. MSTOM is lower than PLTOM thus MSTOM DOM TO Fuel = 62,800 34,930 15,428 = 12,442 kg 3. PLLM is lower than SLLM, thus PLLM - DOM - Fuel Remaining = 54,200 34,930 6,178 = 13,092 kg Allowable TL = lowest of 1, 2 or 3 above = 12,442 kg8. b. The Maximum Structural Landing Mass is set by the manufacturer to meet with the Design Limit Loads (DLL) of the structure. If exceeded, the structure will be subject to excessive fatigue and could even be permanently damaged.9. c. 1. Stick forces at rotation will increase [weight is fwd of wheel rotation] 2. Range will decrease but endurance will increase [ both will decrease] 3. Gliding range will reduce [gliding range is not afected by weight] 4. Stalling speed will increase [stalling speed increases with weight]10. a. Wrong. You will still believe it to be the speed you calculated because you are unaware of the error.b. Correct. VMU will be later (the extra mass will prolong the point of minimum lift-of).c. Wrong. You will pull the stick back to rotate at the speed you originally calculated.d. Wrong. They will all occur later.11 d. the CG is behind the rear limit (review the Principles of Flight notes on Static Margin. If the CG were positioned on the neutral point and the aeroplane was disturbed in pitch by a gust of wind, it would retain the new atitude because the moments about the CG would all equal one another.)12. b. Wrong. The manufacturer sets the limits not the positionc. Correct. within the range set by the manufacturer13. c. alters the moments about the CG but not the CG position14. d. Oceanic means there are no obstacles to consider. Though we have an unlimited runway the high elevation of the airfeld will result in a low air density. Also, the time, being at the hotest part of the day will further reduce the air density. The reduced density will seriously reduce the engine performance limits. Weight would be limited in order to achieve a suitable climb gradient.15. a. 1.3VS = 180 kts, therefore VS = 180 = 138.46 kts

1.3 New VS = Old VS x (New weight/Old Weight) = 138.46 x (320,000/285,000) = 146.7 kts Assuming the aircraft continues to fy at 1.3VS its new speed will be: 146.7 x 1.3 = 190.7 kts Naturally, drag will increase and range (nautical miles per kg of fuel) will decrease).113Chapter 3 Revision Questions16. c The datum does not have to be between the nose and the tail. (The datum an be anywhere in front of, on or behind the aircraft so long as it is on the longitudinal axis of the aeroplane).17. c. TOM = DOM + Trafc Load + Fuel Load. But, Trafc Load + Fuel Load = Useful Load. Thus TOM = DOM + Useful Load, Rearrange formula, UL = TOM DOM18. b. To simplify M&B calculations. See Cap 696 page 419. b. See JAR OPS Sub-part J 1.620, para f20. c. 2, 3, 4 and 5 only In this example the performance limitation is not stated and could be anything from a runway length restriction, a sloping runway, an obstruction limitation and/or altitude/temperature limitations. The aeroplane might sustain a burst tyre, brake fade, and/or brake fre as a result of heavy braking. Tyre temperatures might exceed limits and delay the take-of time even if they dont burst. The climb slope for obstacle clearance during a go-around might be reduced. As the landing is below the MSLM, the structure itself should not sufer direct damage providing the aeroplane comes to a stop without hiting anything.21. d. See CAP 696 Section 1, General Notes, Page 3, Para 4.1.22. b. The pilot calculates the amount of fuel to jetison to reduce the mass to a safe level at or below the RLM. (Before jetisoning the fuel the pilot should atempt to declare an emergency if time permits and advise Air Trafc Control of his intensions).23. a. When atempting this sort of question the golden rule is to work out the fuel states frst. Once the fuel states are known you can simply use the three formulas to determine the answer. TOF = 9700 + 1200 + 1400 = 12300 FR = 1200 + 1400 = 2600 DOM = 34900 MZFM = 53000 RTOM = 62800 RLM = 54900Formulas 1. MZFM DOM = 53,000 34,900 =18,100 2. RTOM DOM TOF = 62,800 -34,900 12,300 =15,600 3. RLM DOM FR = 54,900 -34,900 -2,600 =17,480 Aloowable TL = Lowest of 1, 2, 3 Above = 15,600 kg Acual TL = PAX + Baggage = (130 x 84) + (130 x 14) Actual TL = 12,740 kgDifence Between Allowed TL and Actual TL = Underload.

Cargo That Can Be Taken = Underload Cargo = 15,600 12,740 = 2,860 kg114Revision Questions Chapter 3iiIntroduction Aircraft PerformanceiiiIntroduction Aircraft PerformanceContentsAIRCRAFT PERFORMANCE1. INTRODUCTION. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1152. GENERAL PRINCIPLES - TAKE-OFF . . . . . . . . . . . . . . . . . . . . . . . . . 1333. GENERAL PRINCIPLES - CLIMB AND DESCENT . . . . . . . . . . . . . . . . . 1554. GENERAL PRINCIPLES - CRUISE. . . . . . . . . . . . . . . . . . . . . . . . . . . 2195. GENERAL PRINCIPLES - LANDING . . . . . . . . . . . . . . . . . . . . . . . . . 2596. SINGLE ENGINE CLASS B AIRCRAFT - TAKE OFF . . . . . . . . . . . . . . . . 2797. SINGLE ENGINE CLASS B - CLIMB. . . . . . . . . . . . . . . . . . . . . . . . . . 2938. SINGLE ENGINE CLASS B - EN-ROUTE AND DESCENT . . . . . . . . . . . . . 2999. SINGLE ENGINE CLASS B - LANDING . . . . . . . . . . . . . . . . . . . . . . . 31110. MULTI-ENGINED CLASS B - TAKE OFF . . . . . . . . . . . . . . . . . . . . . . . 32311. MULTI ENGINED CLASS B - EN-ROUTE AND DESCENT. . . . . . . . . . . . . 33512. MULTI-ENGINED CLASS B - LANDING . . . . . . . . . . . . . . . . . . . . . . . 34313. CLASS A AIRCRAFT - TAKE OFF . . . . . . . . . . . . . . . . . . . . . . . . . . . 35314. CLASS A: ADDITIONAL TAKE OFF PROCEDURES . . . . . . . . . . . . . . . . 39315. CLASS A: TAKE OFF CLIMB . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40516. CLASS A: EN ROUTE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42117. CLASS A: LANDING. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44518. REVISION QUESTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455ivIntroduction Aircraft Performance115Chapter 1 IntroductionCHAPTER ONEINTRODUCTIONContentsDEFINITIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117ABBREVIATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125JAR PERFORMANCE CLASSIFICATION . . . . . . . . . . . . . . . . . . . . . . . . 131PERFORMANCE EXPRESSIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131116Chapter 1 Introduction117Chapter 1 IntroductionDEFINITIONSAbsolute Ceiling The altitude at which the theoretical rate of climb, with all engines operating at maximum continuous power, is reduced to zero feet per minute Accelerate Stop Distance Available The distance from the point on the surface of the aerodrome at which the aeroplane can commence its take-of run to the nearest point in the direction of take-of at which the aeroplane cannot roll over the surface of the aerodrome and be brought to rest in an emergency without the risk of accident. It is equal to TORA plus any available stopway. Aerodrome Any area of land or water designed, equipped, set apart or commonly used for afording facilities for the landing and departure of aircraft and includes any area or space, whether on the ground, on the roof of a building or elsewhere, which is designed, equipped or set apart for afording facilities for the landing and departure of aircraft capable of descending or climbing vertically, but shall not include any area the use of which for afording facilities for the landing and departure of aircraft has been abandoned and has not been resumed. Aerodrome Elevation The elevation of the highest point of the landing area. Aerodrome Reference Point The aerodrome reference point is the geographical location of the aerodrome and the centre of its trafc zone where an ATZ is established. Aerodynamic Ceiling The altitude, in unaccelerated 1g level fight, where the Mach number for the low speed and high speed bufet are coincident. Aeroplane A power-driven heavier-than-air aircraft, deriving its lift in fight chiefy from aerodynamic reactions on surfaces which remain fxed under given conditions of fight. Aircraft A machine that can derive support in the atmosphere from the reactions of the air other than the reactions of the air against the earths surface Aircraft Classifcation Number (ACN) This is a value assigned to an aeroplane to show its load force. The aircraft classifcation number must be compared to the pavement classifcation number (PCN) of an aerodrome. The aircraft classifcation number may exceed the pavement classifcation number by as much as 50% but only if the maneuvering of the aeroplane is very carefully monitored otherwise signifcant damage may occur to both the aeroplane and the pavement.Airframe The fuselage, booms, nacelles, cowlings, fairings, aerofoil surfaces (including rotors but excluding propellers and rotating aerofoils of engines), and landing gear of an aircraft and their accessories and controls Air Minimum Control Speed The minimum speed at which directional control can be demonstrated when airborne with the critical engine inoperative and the remaining engines at take-of thrust. Full opposite rudder and not more than 5 degrees of bank away from the inoperative engine are permited when establishing this speed. VMCA may not exceed 1.2 Vs. Alternate airport means an airport at which an aircraft may land if a landing at the intended airport becomes inadvisable.Altitude The altitude shown on the charts is pressure altitude. This is the height in the International Standard Atmosphere at which the prevailing pressure occurs. It may be obtained by seting the sub-scale of a pressure altimeter to 1013 hPa. 118Chapter 1 IntroductionAngle of atack The angle between the chord line of the wing of an aircraft and the relative airfow. Apron A defned area on a land aerodrome provided for the stationing of aircraft for the embarkation and disembarkation of passengers, the loading and unloading of cargo, and for parking. Auxiliary Power Unit Any gas turbine-powered unit delivering rotating shaft power, compressor air, or both which is not intended for direct propulsion of an aircraft Balanced Field A runway for which the Accelerate Stop Distance Available is equal to the Take-of Distance Available is considered to have a balanced feld length. Baulked Landing A landing manoeuvre that is unexpectedly discontinued Brake Horsepower The power delivered at the main output shaft of an aircraft engine Bufet Speed The speed at which the airfow over the wing separates creating turbulent airfow aft of the separation point which bufets the aeroplane Calibrated Airspeed The indicated air speed, corrected for position and instrument error. It is equal to True Airspeed (TAS) at Mean Sea Level (MSL) in a Standard Atmosphere Climb Gradient The ratio, in the same units of measurement, expressed as a percentage, as obtained from the formula:- Gradient = Vertical interval the horizontal interval x 100. Clearway An area beyond the runway, not less than 152 m (500 ft) wide, centrally located about the extended centreline of the runway, and under the control of the airport authorities. The clearway is expressed in terms of a clearway plane, extending from the end of the runway with an upward slope not exceeding 125%, above which no object or terrain protrudes. However, threshold lights may protrude above the plane if their height above the end of the runway is 066 m (26 inches) or less and if they are located to each side of the runway.Cloud Ceiling In relation to an aerodrome, cloud ceiling means the vertical distance from the elevation of the aerodrome to the lowest part of any cloud visible from the aerodrome which is sufcient to obscure more than one half of the sky so visible. Contaminated runway A runway covered by more than 25% of water whose equivalent depth is 3mm. Continuous One Engine Inoperative Power Rating The minimum test bed acceptance power, as stated in the engine type certifcate data sheet, when running at the specifed conditions and within the appropriate acceptance limitations Continuous One Engine Inoperative Thrust Rating The minimum test bed acceptance thrust, as stated in the engine type certifcate data sheet, when running at the specifed conditions and within the appropriate acceptance limitations Continuous One Engine Inoperative Power The power identifed in the performance data for use after take-of when a power-unit has failed or been shut down, during periods of unrestricted duration Continuous One Engine Inoperative Thrust The thrust identifed in the performance data for use after take-of when a power-unit has failed or been shut down, during periods of unrestricted duration 119Chapter 1 IntroductionCritical Engine The engine whose failure would most adversely afect the performance or handling qualities of an aircraft Damp runway A runway is considered damp when the surface is not dry, but when the moisture on it does not give it a shiny appearance. Declared Distances The distances declared by the aerodrome authority for the purpose of application of the requirement of the Air Navigation Decision Speed The maximum speed in the take-of at which the pilot must take the frst action (e.g. apply brakes, reduce thrust, deploy speed brakes) to stop the aeroplane within the accelerate-stop distance. It also means the minimum speed in the take-of, following a failure of the critical engine at VEF, at which the pilot can continue the take-of and achieve the required height above the take-of surface within the takeof distance Density Altitude The altitude in I.S.A., where the prevailing measured density occurs. Drag That force on an aeroplane which directly opposes thrust Dry runway A dry runway is one which is neither wet nor contaminated, and includes those paved runways which have been specially prepared with grooves or porous pavement and maintained to retain efectively dry braking action even when moisture is present. Elevation The vertical distance of an object above mean sea level. This may be given in metres or feet. En-route The en-route phase extends from 1500 ft above the take-of surface level to 1000 ft above the landing aerodrome surface level for class B aeroplanes or to 1500 ft above the landing aerodrome surface level for class A aeroplanes. Equivalent Airspeed The calibrated airspeed corrected for compressibility at the particular pressure altitude under consideration. It is equal to Calibrated Airspeed in a Standard Atmosphere Exhaust Gas Temperature The average temperature of the exhaust gas stream Final en-route climb speed The speed of the aeroplane in segment four of the take-of fight path with one engine inoperative Final Segment Speed The speed of the aeroplane in segment four of the take-of fight path with one engine inoperative Final take-of speed The speed of the aeroplane that exists at the end of the take-of path in the en-route confguration with one engine inoperative Fixed Pitch Propeller A propeller, the pitch of which cannot be changed, except by processes constituting a workshop operation Flap extended speed The highest speed permissible with wing-faps in a prescribed extended position Flight Level A surface of constant atmospheric pressure that is related to 1013.25 hPa. It is conventionally the pressure altitude to the nearest 1000 ft in units of 100 ft. For example, fight level 250 represents a pressure altitude of 25,000 feet. 120Chapter 1 IntroductionFrangibility The ability of an object to retain its structural integrity and stifness up to a specifed maximum load but when subject to a load greater than specifed or struck by an aircraft will break, distort or yield in such a manner as to present minimum hazard to an aircraft. Go-around A procedure involving a decision to abort the landing and climb straight ahead to rejoin the circuit. Such a decision might be taken at any time during the fnal approach, the transition phase or even after initial touchdown. Gross Height The true height atained at any point in the take-of fight path using gross climb performance. Gross height is used for calculating pressure altitudes for purposes of obstacle clearance and the height at which wing fap retraction is initiated. Gross Performance The average performance that a feet of aeroplanes should achieve if satisfactorily maintained and fown in accordance with the techniques described in the manual. Ground Minimum Control Speed The minimum speed at which the airplane can be demonstrated to be controlled on the ground using only the primary fight controls when the most critical engine is suddenly made inoperative and the remaining engines at take-of thrust. Throtling an opposite engine is not allowed in this demonstration. Forward pressure from the elevators is allowed to hold the nose wheel on the runway, however, nose wheel steering is not allowed. Height The vertical distance between the lowest part of the aeroplane and the relevant datum. Hydroplaning Speed The speed at which the wheel is held of the runway by a depth of water and directional control through the wheel is impossible. ICAO Standard Atmosphere The atmosphere defned in ICAO Document 7488/2. For the purposes of Certifcation Specifcations the following is acceptable: The air is a perfect dry gas The temperature at sea-level is 15C The pressure at sea-level is 10132 hPa (2992 in Hg) The temperature gradient from sea-level to the altitude at which the temperature becomes 565C is 0.65C/100 m (198C/1 000 ft) The density at sea level under the above conditions is 12250 kg/m3

IFR conditions Weather conditions below the minimum for fight under visual fight rules Indicated Airspeed The speed as shown by the pitot/static airspeed indicator calibrated to refect Standard Atmosphere adiabatic compressible fow at MSL and uncorrected for airspeed system errors Instrument A device using an internal mechanism to show visually or aurally the atitude, altitude, or operation of an aircraft or aircraft part. It includes electronic devices for automatically controlling an aircraft in fight Landing Distance Available The distance from the point on the surface of the aerodrome above which the aeroplane can commence its landing, having regard to the obstructions in its approach path, to the nearest point in the direction of landing at which the surface of the aerodrome is incapable of bearing the weight of the aeroplane under normal operating conditions or at which there is an obstacle capable of afecting the safety of the aeroplane. 121Chapter 1 IntroductionLanding gear extended speed The maximum speed at which an aircraft can be safely fown with the landing gear extended Landing gear operating speed The maximum speed at which the landing gear can be safely extended or retracted Landing Minimum Control Speed The minimum speed with a wing engine inoperative where it is possible to decrease thrust to idle or increase thrust to maximum take-of without encountering dangerous fight characteristics. Large aeroplane An aeroplane of more than 5,700 kg maximum certifcated take-of weight. The category Large Aeroplane does not include the commuter aeroplane category Lift That force acting on an aerofoil which is at right angles to the direction of the airfow Load factor The ratio of a specifed load to the total weight of the aircraft. The specifed load is expressed in terms of any of the following: aerodynamic forces, inertia forces, or ground or water reactionsMach number The ratio of true air speed to the Local Speed of Sound (LSS)Manoeuvre Ceiling The altitude, in unaccelerated 1.3g level fight, where the mach number for the low speed and high speed bufet are coincident. Maximum Brake Energy Speed The maximum speed on the ground from which an aeroplane can safely stop within the energy capabilities of the brakes. Maximum Continuous Power The power identifed in the performance data for use during periods of unrestricted duration Maximum Continuous Thrust The thrust identifed in the performance data for use during periods of unrestricted duration Maximum Structural Take-Of Mass The maximum permissible total mass of an aeroplane at the start of the take-of run. Maximum Structural Landing Mass The maximum permissible total mass of an aeroplane on landing (under normal circumstances). Minimum Control speed The minimum speed at which the airplane is directionally controllable with the critical engine inoperative and the remaining engines at take-of thrust. Minimum unstick speed The minimum speed demonstrated for each combination of weight, thrust, and confguration at which a safe takeof has been demonstrated.Missed Approach When an aircraft is caused to abort a landing after it has already started its landing approach. The aircraft has to follow a set missed approach procedure to leave the airspace surrounding the terminal. Net Height The true height atained at any point in the take-of fight path using net climb performance. Net height is used to determine the net fight path that must clear all obstacles by the statutory minimum to comply with the Operating Regulations. 122Chapter 1 IntroductionNet Performance Net performance is the gross performance diminished to allow for various contingencies that cannot be accounted for operationally e.g., variations in piloting technique, temporary below average performance, etc. It is improbable that the net performance will not be achieved in operation, provided the aeroplane is fown in accordance with the recommended techniques. Outside Air Temperature The free air static (ambient) temperature. Pitch Seting The propeller blade seting determined by the blade angle, measured in a manner and at a radius declared by the manufacturer and specifed in the appropriate Engine Manual Pitch One of the three axis in fight, this specifes the up and down movement of the aircraft about its lateral axis. Pitot tube A small tube most often mounted on the outward leading edge of an airplane wing (out of the propeller stream) that measures the dynamic pressure of the air it meets in fight, working in conjunction with a closed, perforated, coaxial tube that measures the static pressure. The diference in pressures is calibrated as airspeed. Pressure Altitude The altitude of an aircraft above the pressure level of 1013.25 hPa. This is achieved by seting the altimeter sub-scale to 1013 hPa and reading the altitude indicated. Reference landing speed The speed of the aeroplane, in a specifed landing confguration, at the point where it descends through the landing screen height in the determination of the landing distance for manual landings Roll One of the three axis in fight, this specifes the rolling movement of the aeroplane about its longitudinal axis. Rotation Speed The speed at which, during the take-of, rotation is initiated with the intention of becoming airborne. Runway A defned rectangular area, on a land aerodrome prepared for the landing and take-of run of aircraft along its length. Runway Strip An area of specifed dimensions enclosing a runway intended to reduce the risk of damage to an aircraft running of the runway and to protect aircraft fying over it when taking-of or landing. Runway Threshold The beginning of that portion of the runway usable for landing. Screen Height The height of an imaginary screen placed at the end of the Take-Of Distance Required and at the beginning of the Landing Distance Required. Service Ceiling The altitude at which the theoretical rate of climb becomes less than a specifed rate of climb (typically 1000 ft/min or 300 ft/min), with all engines operating at maximum continuous power. Specifc Fuel Consumption Fuel fow per unit thrust. The lower the value, the more efcient the engine. Stopway An area beyond the take-of runway, no less wide than the runway and centred upon the extended centreline of the runway, able to support the aeroplane during an abortive take-of, without causing structural damage to the aeroplane, and designated by the airport authorities for use in decelerating the aeroplane during an abortive take-of.123Chapter 1 IntroductionTake-of Distance Available. The distance from the point on the surface of the aerodrome at which the aeroplane can commence its take-of run to the nearest obstacle in the direction of take-of projecting above the surface of the aerodrome and capable of afecting the safety of the aeroplane. It is equal to TORA plus any clearway and cannot be more than one and one half times the TORA, whichever is the less. Take-Of Mass The mass of an aeroplane, including everything and everyone contained within it, at the start of the take-of run. Take-of Power The output shaft power identifed in the performance data for use during take-of, discontinued approach and baulked landing;i. for piston engines, it is limited in use to a continuous period of not more than 5 minutes;ii. for turbine engines installed in aeroplanes and helicopters, limited in use to a continuous period of not more than 5 minutes; andiii. for turbine engines installed in aeroplanes only (when specifcally requested), limited in use to a continuous period of not more than 10 minutes in the event of a power-unit having failed or been shut down Take-of Run Available The distance from the point on the surface of the aerodrome at which the aeroplane can commence its take-of run to the nearest point in the direction of take-of at which the surface of the aerodrome is incapable of bearing the weight of the aeroplane under normal operating conditions. Take-of safety speed A referenced airspeed obtained after lift-of at which the required one engine-inoperative climb performance can be achieved Take-of Thrust means the output shaft thrust identifed in the performance data for use during take-of, discontinued approach and baulked landing;i. for piston engines, it is limited in use to a continuous period of not more than 5 minutes;ii. for turbine engines installed in aeroplanes and helicopters, limited in use to a continuous period of not more than 5 minutes; andiii. for turbine engines installed in aeroplanes only, limited in use to a continuous period of not more than 10 minutes in the event of a power-unit having failed or been shut down Taxiway A defned path on a land aerodrome established for the taxiing of aircraft and intended to provide a link between one part of the aerodrome and another Thrust That force acting on an aeroplane produced by the engine(S) in a forward direction. True airspeed The airspeed of an aircraft relative to undisturbed air Turbo jet An aircraft having a jet engine in which the energy of the jet operates a turbine that in turn operates the air compressor. Turbo prop An aircraft having a jet engine in which the energy of the jet operates a turbine that drives the propeller. Turboprops are often used on regional and business aircraft because of their relative efciency at speeds slower than, and altitudes lower than, those of a typical jet. Variable Pitch Propellers A propeller, the pitch seting of which changes or can be changed, when the propeller is rotating or stationary 124Chapter 1 IntroductionVEF The calibrated airspeed at which the critical engine is assumed to fail and is used for the purpose of performance calculations. It is never less than VMCG.V1 Referred to as the decision speed. It is the speed at which the pilot, in the event of a power unit failure, must decide whether to abandon or continue the take-of. It is the maximum speed in the take-of at which the pilot must take the frst action (e.g. apply brakes, reduce thrust, deploy speed brakes) to stop the aeroplane within the accelerate-stop distance. It also means the minimum speed in the take-of, following a failure of the critical engine at VEF, at which the pilot can continue the take-of and achieve the required height above the take-of surface within the takeof distance. Engine failure prior to V1 demands that the pilot must abandon the take-of because there is insufcient distance remaining to enable the aircraft accelerate to the screen height. Engine failure after V1 demands that the pilot must continue the take-of because there is insufcient distance remaining to safely bring the aircraft to a stop Wet runway A runway is considered wet when the runway surface is covered with water, or equivalent moisture on the runway surface to cause it to appear refective, but without signifcant areas of standing water. Windshear Localised change in wind speed and/or direction over a short distance, resulting in a tearing or shearing efect that can cause a sudden change of airspeed with occasionally disastrous results if encountered when taking-of or landing. Yaw One of the three axis in fight, this specifes the side-to-side movement of an aircraft about its vertical axis. Zero fap speed The minimum safe maneuvering speed with zero fap selected 125Chapter 1 IntroductionABBREVIATIONSAC Air Conditioning ACARS Aircraft Communications Addressing and Reporting System ACS Air Conditioning System ACN Aircraft Classifcation Number AFM Aeroplane Flight Manual AGL Above ground level AMSL Above mean sea level ANO Air Navigation Order AOM Airline Operation Manual ASD Accelerate Stop Distance ASDA Accelerate Stop Distance Available ASDR Accelerate Stop Distance Required ATC Air Trafc Control AUW All-Up Weight BRP Brake Release Point CAA Civil Aviation Authority CAP Civil Aviation Publication CAS Calibrated Airspeed CI Cost Index CL Lift coefcient C of A Certifcate of Airworthiness C of G Centre of Gravity C of P Centre of Pressure FMC Flight Management Computer FMS Flight Management System DL Drag coefcient DOC Direct Operating Cost 126Chapter 1 IntroductionDOW Dry operating weight EAS Equivalent Air Speed EASA European Aviation Safety Agency ECON Economic speed (minimum directing operating cost speed) EGT Exhaust Gas Temperature EMD Emergency Distance EMDA Emergency Distance Available EMDR Emergency Distance Required ETOPS Extended range with twin aeroplane operations FAA Federal Aviation Administration FCOM Flight Crew Operating Manual FF Fuel Flow (hourly consumption) FL Flight Level G/S Ground Speed hPa Hectopascal IAS Indicated Air Speed IAT Indicated Air Temperature ICAO International Civil Aviation Organisation IFR Instrument fight rules ILS Instrument Landing System ISA ICAO Standard Atmosphere JAA Joint Aviation Authority JAR Joint Aviation Requirements kg Kilograms km Kilometres kt Nautical miles per hour (knots) LCN Load Classifcation Number LDA Landing Distance Available 127Chapter 1 IntroductionLDR Landing Distance Required LRC Long Range Cruise speed MAT Mass-altitude-temperature MCRT Critical Mach number MCT Maximum Continuous Thrust MEA Minimum safe En route Altitude MEL Minimum Equipment List MMO Maximum Operating Mach number MMR Mach of Maximum Range MSL Mean Sea Level MTOW Maximum Take-Of Weight MLRC Mach Number for Long Range MSL Mean Sea Level MZFW Maximum Zero Fuel Weight N All engines operating. NFP Net Flight Path nm nautical mile NOTAM Notice to Airmen N1 Speed rotation of the fan N-1 One engine inoperative N-2 Two engines inoperative OAT Outside Air Temperature OEI One engine inoperative

PA Pressure Altitude PCN Pavement Classifcation Number PFD Primary Flight Display PMC Power Management ControlPNR Point of No Return 128Chapter 1 Introductionpsi Pounds per square inch QFE The altimeter sub-scale seting which causes the altimeter to read zero elevation when on the airfeld reference point or runway threshold QNE The indicated height on the altimeter at the aerodrome datum point with the altimeter sub-scale set to 1013.2 hPa QNH The altimeter sub-scale seting which causes the altimeter to read the elevation of the airfeld above mean sea level when placed on the airfeld reference point or runway threshold RESA Runway End Safety Area RNP Required Navigation Performance RZ Reference Zero SAR Specifc Air Range SFC Specifc Fuel Consumption SR Specifc Range TAS True Airspeed TAT Total Air Temperature TOD Take-Of Distance/Top of Descent TODA Take-Of Distance Available TODR Take-Of Distance Required TOGA Take-of/Go-around thrust TOR Take-Of Run TORA Take-Of Run Available TORR Take-Of Run Required TOSS Take-Of Safety Speed TOW Take-Of Weight V1 Decision Speed V2 Take-Of Safety Speed V2MIN Minimum take-of safety speed V3 All-engines operating steady initial climb speed V4 All-engines operating steady take-of climb speed 129Chapter 1 IntroductionVA Design Maneuvering Speed VEF The assumed speed of engine failure VFE The maximum fap extended speed VFR Visual fight rules VFTO Final take-of speedVGO The lowest decision speed from which a continued take-of is possible with TODA with one engine inoperative VMD Velocity of Minimum Drag VMP Velocity of Minimum Power VLE The maximum speed with landing gear extended VLO The maximum at which the landing gear may be lowered VLOF Lift-of Speed VMBE Maximum brake-energy speed VMC Minimum control speed with the critical power unit inoperative VMCA Air minimum control speed in the air (take-of climb) VMCG Ground minimum control speed (at or near the ground) VMCL Landing minimum control speed (on the approach to land) VMO The maximum operating speed VMU The minimum unstick speed VNE Never exceed speed VP Hydroplaning/Aquaplaning speed VR Rotation Speed VRA The turbulence speed or rough air speed VREF The reference landing speed (Replaced VAT speed) VS Stalling speed or minimum steady fight speed at which the aeroplane is controllable VSR Reference stalling speed. Assumed to tbe same as VS1G VSR0 Reference stalling speed in the landing confguration VSR1 Reference stalling speed in the specifed confguration 130Chapter 1 IntroductionVS1G Stalling speed at 1g or the one-g stall speed at which the aeroplane can develop a lift force (normal to the fight path) equal to its weight. This is assumed to be the same speed as VSR. VS0 The stalling speed with the faps at the landing seting or minimum steady fight speed at which the aeroplane is controllable in the landing confguration VS1 The stalling speed for the confguration under consideration VSTOP The highest decision speed that an aeroplane can stop within ASDA VTIRE Maximum tire speed VX The speed for the best gradient or angle of climb VY The speed for the best rate of climb VZF The minimum safe maneuvering speed with zero fap WAT Weight-altitude-temperature ZFW/ZFM Zero fuel weight/zero fuel mass Climb or descent angle Runway friction coefcient Aircraft atitude Air density 131Chapter 1 IntroductionJAR PERFORMANCE CLASSIFICATIONPERFORMANCE CLASS AMulti-engined aeroplanes powered by turbo-propeller engines with a maximum approved passenger seating confguration of more than 9 or a maximum take-of mass exceeding 5700 kg., and all multi-engined turbo-jet powered aeroplanes. Class A aeroplanes must abide by the Certifcation Specifcations laid out in the document from EASA called CS-25.PERFORMANCE CLASS BPropeller driven aeroplanes with a maximum approved passenger seating confguration of 9 or less, and a maximum take-of mass of 5700 kg. or less. Class B aeroplanes must abide by the Certifcation Specifcations laid out in the document from EASA called CS-23.PERFORMANCE CLASS CAeroplanes powered by reciprocating engines with a maximum approved passenger seating confguration of more than 9 or a maximum take-of mass exceeding 5700 kg.UNCLASSIFIEDThis class is given to those aeroplanes whose performance characteristic is very unique and special performance consideration is required. For example, the Unclassifed class includes supersonic aeroplanes and sea planes.Multi-engined JetPropeller drivenMulti-engined Turbo-propPistonMass : > 5700 kgorPassenger Seats: > 9A A CMass : 5700 kgandPassengers Seats : 9A B BPERFORMANCE EXPRESSIONS Any class of aeroplane, operated in the public transport role must adhere to the operational requirements set out in JAR-OPS 1. JAR-OPS 1 prescribes a minimum performance level for each stage of fight for Class A, Class B and Class C aeroplanes. The certifcation and operational regulations together aim to achieve a high standard of safety that has kept air travel as the safest form of travel. To achieve the required safety standard the aviation authorities have added a safety margin into the aeroplane performance data. The application of these safety margins changes the expression of the performance data.132Chapter 1 IntroductionMEASURED PERFORMANCEThis is the performance achieved by the manufacturer under test conditions for certifcation. It utilizes new aeroplanes and test pilots and it therefore unrepresentative of the performance that will be achieved by an average feet of aeroplanes.GROSS PERFORMANCEGross Performance is the average performance that a feet of aeroplanes should achieve if satisfactorily maintained and fown in accordance with the techniques described in the manual. Therefore, Gross Performance is Measured Performance reduced by a set margin to refect average operating performance.NET PERFORMANCENet performance is the gross performance diminished to allow for various contingencies that cannot be accounted for operationally e.g., variations in piloting technique, temporary below average performance, etc. It is improbable that the net performance will not be achieved in operation, provided the aeroplane is fown in accordance with the recommended techniques. This level of performance is approximately 5 standard deviations from the average performance or gross performance. Therefore, 99.99994% of the time, the aeroplane will achieve net performance or beter. However, there is less than one chance in a million that the aeroplane will not achieve the net performance. This is the safety standard which the aviation authorities aim to achieve.133Chapter 2 General Principles -Take-OffCHAPTER TWOGENERAL PRINCIPLES - TAKE-OFFContentsTAKE OFF. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135THE TAKE-OFF RUN AVAILABLE (TORA) . . . . . . . . . . . . . . . . . . . . . . . 137FORCES DURING TAKE-OFF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139SUMMARY OF FORCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143EFFECT OF VARIABLE FACTORS ON TAKE OFF DISTANCE . . . . . . . . . . . . 144QUESTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149ANSWERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154134Chapter 2 General Principles -Take-Off135Chapter 2 General Principles -Take-OffTAKE OFFThe take of part of the fight is the distance from the brake release point (BRP) to the point at which the aircraft reaches a defned height. This defned height is termed the screen height. The screen height varies from 35 ft for class A aeroplanes to 50 ft for class B aeroplanes.For any particular take-of, it must be shown that the distance required for take-of in the prevailing conditions does not exceed the distance available at the take of aerodrome. However, there are various terms used to describe the available distances at an aerodrome.AVAILABLE DISTANCESMost large aerodromes have extra distances associated with the main runway which are used in a variety of ways. These extra distances are Clearways and Stopways.CLEARWAYSClearways are an area beyond the runway, not less than 152 m (500 ft) wide, centrally located about the extended centreline of the runway, and under the control of the airport authorities. The clearway is expressed in terms of a clearway plane, extending from the end of the runway with an upward slope not exceeding 125%, above which no object or terrain protrudes. However, threshold lights may protrude above the plane if their height above the end of the runway is 066 m (26 inches) or less and if they are located to each side of the runway. Clearways are not physical structures; they are simply an area of defned width and length which are free of obstacles.

Figure 2.1 Clearways extend from the end of the runway with an upward slopenot exceeding 125%, above which no object or terrain protrudes.136Chapter 2 General Principles -Take-OffSTOPWAYSStopways are an area beyond the take-of runway, no less wide than the runway and centred upon the extended centreline of the runway, able to support the aeroplane during an abortive take-of, without causing structural damage to the aeroplane, and designated by the airport authorities for use in decelerating the aeroplane during an abortive take-of. Stopways are physical structures and are usually paved. However, stopways are not as strong as the main length of runway and therefore are only used to help bring an aeroplane to a stop in the event of an abandoned take-of. Stopways are identifed by large yellow chevrons on either end of the main runway.Figure 2.2 Stopways are able to support the aeroplane during an abortive take-of and are marked by large yellow chevrons.Understanding Stopways and Clearways is essential when examining the available distances published at aerodromes. There are four principle published aerodrome distances, although only three will apply to the take-of and these are discussed next.Figure 2.3 Illustrates the Stopways and Clearways that can be found at aerodromes.137Chapter 2 General Principles -Take-OffTHE TAKE-OFF RUN AVAILABLE (TORA)The take-of run available is the distance from the point on the surface of the aerodrome at which the aeroplane can commence its take-of run to the nearest point in the direction of take-of at which the surface of the aerodrome is incapable of bearing the weight of the aeroplane under normal operating conditions. At most aerodromes the take-of run available is the length of the runway from threshold to threshold.THE TAKE-OFF DISTANCE AVAILABLE (TODA)The take-of distance available is the take-of run available plus any clearway (TORA + Clearway). If there is no clearway at the aerodrome then the take-of distance available will be the same length as the take-of run available. The take-of distance available must be compared to the aeroplanes actual take-of distance. The requirements for take-of state that the aeroplane must be able to complete the take-ofwithin the take-of distance available. Although clearways can be of any length, there is a limit to the amount of clearway that can be used when calculating the TODA. The maximum length of clearway is this case cannot be more than half the length of TORA.THE ACCELERATE-STOP DISTANCE AVAILABLE OR EMERGENCY DISTANCE AVAILABLE (ASDA/EMDA)The accelerate-stop distance available is the length of take-of run available plus any stopway (TORA + Stopway). If there is no stopway at the aerodrome then the accelerate-stop distance available will be the same length as the take-of run available.The accelerate-stop distance available must be compared to the aeroplanes actual accelerate stop distance. The requirements for take-of state that the aeroplanes accelerate-stop distance must not exceed the accelerate-stop distance available.Figure 2.1Figure 2.4 A summary of the available distances at an aerodrome.138Chapter 2 General Principles -Take-OffREQUIRED DISTANCESThe distance required for take-of may be considered as two segments: The take of roll or ground run. The airborne distance to a "screen" of defned height.The total distance from the brake release point to the screen height is called the Take-of Distance (TOD). The speed at which the pilot will atempt to raise the nose wheel of the ground is called VR. This is the speed for rotation. At this speed the pilot will pull back on the control column and eventually the nose wheel will lift of the runway. This action will increase the lift and eventually the main wheels themselves will lift of the runway. The speed at which this occurs is called VLOF which means the speed for lift of.Figure 2.5. The Take-Of Distance (TOD) is the total distance frombrake release until the screen height.Calculating the take-of distance is an important consideration but it is essential to a pilot to understand that there is a chance, albeit a remote chance, that the actual calculated take-ofdistance will not be achieved. To this end, the aviation authorities have created a set of safety regulations to ensure that poor performance is accounted for in the calculations. These safety regulations require factors to be applied to the estimated distances to give a satisfactory safety margin. The application of safety factors changes the terminology used. The calculated take-of distance is called the Gross Take-of distance, but applying the safety factors changes this to the Net Take-of Distance. The regulations and their associated safety factors will be discussed later in the relevant chapters on the diferent classes of aeroplanes.139Chapter 2 General Principles -Take-OffCALCULATING THE TAKE-OFF DISTANCE.The two formulas shown below are used to help understand the key components in calculating the take-of distance. The upper formula is used to calculate the distance required (s) to reach a specifed speed (V) with a given acceleration (). Beneath the frst formula is the formula to calculate that acceleration. s =

v2

2aFigure 2.6 A simplifed take-of distance and acceleration formula.For an aircraft taking of, the acceleration is thrust minus drag. However, both of these forces will change as the speed changes, and so the acceleration will not be constant during the take of. Also, during the airborne part of the take-of, the laws of motion will be somewhat diferent, and the upper formula will change somewhat, but the distance required will still depend on the speed to be achieved and the acceleration.FORCES DURING TAKE-OFFFigure 2.3.Figure 2.7 Forces in the take-of.Looking back at the formula for acceleration during take-of it can see that thrust and drag play a very crucial part in the take-of performance. Therefore a litle extra detail will be covered on these two important forces.140Chapter 2 General Principles -Take-OffTHRUSTThe engine thrust will vary during take of, and the variation of thrust with speed will be diferent for jet and propeller engines. Jet engine. For a jet engine the net thrust is the diference between the gross thrust and the intake momentum drag. Increasing speed increases the intake momentum drag, which reduces the thrust. However, at higher speeds the increased intake pressure due to ram helps to reduce this loss of thrust, and eventually at very high speeds it will cause the net thrust to increase again. During take-of the aeroplane speed is still low and as such the ram efect is insufcient to counteract the loss of thrust due to intake momentum drag, therefore during the take-of there will be a decrease of thrust. In later chapters and in some performance graphs you will notice that the assumption is made that jet thrust is constant with speed. This is done so as to simplify some of the teaching points.Figure 2.8 Net Thrust with speed for a typical modern jet engine. Flat rated engines. The thrust produced by an engine at a given rpm. will depend on the air density, and hence on air pressure and temperature. At a given pressure altitude, decreasing temperature will give increasing thrust. However, many jet engines are fat rated, that is, they are restricted to a maximum thrust even though the engine is capable of producing higher thrust. The reason being that at lower temperatures too much thrust may be generated and the pressures within the compressors may be exceeded. Consequently at temperatures below the fat rating cut of, (typically about ISA + 15C) engine thrust is not afected by temperature. 141Chapter 2 General Principles -Take-OffFigure 2.5Figure 2.9 Typical thrust with air temperature from a fat rated engine. Propeller. For a propeller driven aircraft, thrust is produced by a propeller converting the shaft torque into propulsive force. For a fxed pitch propeller, angle of atack decreases as forward speed increases. Thrust therefore decreases with increasing speed. For a variable pitch propeller, the propeller will initially be held in the fne pitch position during take of and the propeller angle of atack will decrease with increasing speed. Above the selected rpm. the propeller governor will come into operation, increasing the propeller pitch, and reducing the rate at which the thrust decreases. In summary therefore, the thrust of a propeller aeroplane decreases with forward speed.Figure 2.0 Typical thrust with air temperature from a fat rated engine.Supercharged engines. If the engine is un-supercharged, the power produced will decrease with decreasing density, (higher temperature or lower pressure). For a supercharged engine, power may be maintained with increasing altitude, although increasing temperature will cause a decrease in density and a loss of power.142Chapter 2 General Principles -Take-OffDRAGThe total drag (D) of an aeroplane during take-of is a product of both aerodynamic drag (DA) and wheel drag/wheel friction () as shown in the formula below. D = DA + (W - L) Aerodynamic Drag. There are principally two forms of aerodynamic drag, parasite drag and induced drag. Parasite drag is increases by the square of the speed, therefore this form of drag will increase during the take-of. Induced drag is a function of the angle of atack. This angle of atack is constant until the aeroplane rotates at which point the angle of atack increases dramatically. Therefore induced drag will increase during the take-of. Wheel Drag. The wheel drag depends on the load on the wheel (W-L) and the runway surface resistance (). At the start of the take-of the load on the wheels is the entire weight of the aeroplane, therefore wheel friction and wheel drag is high. However, as forward speed increases, lift starts to counteract the weight force and this reduces the load on the wheels. Therefore the wheel friction and the wheel drag will reduce, eventually being zero at lift of.The increase of aerodynamic drag is much higher than the decrease of wheel drag, therefore total drag during the take-of increases.143Chapter 2 General Principles -Take-OffSUMMARY OF FORCESIn summary then, for all aeroplanes during the take-of, thrust decreases and drag increases. The acceleration force is determined by subtracting the total drag from the total thrust. This is visually represented in the next graph by the area between the total drag line and the thrust line. When thrust is more than drag a more commonly used term Excess Thrust is used. Excess Thrust is what is needed to accelerate the aeroplane. You can see from the graph that the Excess thrust and therefore the acceleration of the aeroplane decreases during the take-of. The term Excess Thrust will be used again when Climb theory is introduced.Figure 2.1 Variation of forces with speed during the take-ofTAKE OFF SPEEDThe speed (V) in the take-of distance formula is True Ground Speed. When calculating the take-of run required, account must therefore be taken of the efect of density on TAS for a given IAS, and of the efect of wind on TGS for a given TAS.The speed to be reached at the screen (the Take-of Safety Speed) is determined by the Regulations, and is required to be a safe margin above the stall speed and the minimum control speed, a speed that gives adequate climb performance, and that takes account of the acceleration that will occur after lift of. It is very important to ensure this speed is achieved by the screen height. 144Chapter 2 General Principles -Take-OffEFFECT OF VARIABLE FACTORS ON TAKE OFF DISTANCEMASS The mass of the aeroplane afects:The acceleration for a given accelerating force. This is the efect of inertia. An aeroplane with higher mass will have more inertia. Therefore as mass increases, acceleration will decrease which will increase the take-of distance. The wheel drag. Increased mass increases the load placed on the wheels and therefore increases the wheel friction. Because of the increased wheel friction, wheel drag will increase. Therefore, acceleration is reduced and the take-of distance will increase.The take of safety speed. An aeroplane with a higher mass will have a greater force of weight. This must be overcome by greater lift. To gain this extra lift the aeroplane must be accelerated to a higher speed, this will of course increase the take-of distance.The angle of initial climb to the screen height. This efect will be beter understood in the next chapter, but nonetheless a higher mass reduces the angle of the initial climb. This means that the aeroplane will use a greater horizontal distance to get to the screen height.In summary then, increasing mass has four detrimental efects to the take-of distance.AIR DENSITYDensity is determined by pressure, temperature and humidity. Density afects: The power or thrust of the engine. Reduced density will reduce combustion inside the engine and therefore reduce the thrust and or power that the engine can generate. Therefore acceleration will be less and the take-of distance will increase. The TAS for a given IAS. Reduced density will increase the true airspeed for a given indicated airspeed. For example, if the take-of safety speed was an indicated airspeed of 120 knots, then in low density this may represent a true airspeed of 130 knots. Geting to a true speed of 130 knots will require more distance. Therefore, low density will increase the take-of distance. The angle of the initial climb. Since there is less thrust and or power in low density, the angle of climb will reduce. Therefore, geting to the screen height will require a longer horizontal distance.WINDWinds afect the true ground speed of the aeroplane for any given true airspeed. Headwinds will reduce the ground speed at the required take-of air speed and reduce the take of distance. For example, with a headwind of 20 knots and a true airspeed for the take-of safety speed being 120 knots, the ground speed is only 100 knots. Geting to a true ground speed of only 100 knots will require less distance. Another beneft is that headwinds also increase the angle of the initial climb which will further reduce the required distance. Therefore, headwinds reduce the take-of distance and it is this reason why pilots always aim to take-of into wind.A tailwind does the opposite to a headwind. Tailwinds will increase the ground speed and increase the take of distance.The Regulations for all classes of aircraft require that in calculating the take-of distance, no more than 50% of the headwind component is assumed and no less than 150% of a tailwind component is assumed. This is to allow for variations in the reported winds during take of. 145Chapter 2 General Principles -Take-OffFor example, it would not be wise to plan a maximum take-of mass with 10 knots headwind if at the actual time of take-of the wind was less than 10 knots. In this case, the aeroplane would not be able to complete the take-of within the available distance. Most aeroplane performance manuals and operating handbooks already have the wind rules factored into the take-of graphs or tables. In this case, simply used the forecast wind and the graph or table will automatically correct the take-of distance to account for the regulation on wind.Note: For any headwind the distance required to take-of will be less than the calculated distance, as only half the headwind is allowed for. Equally for any tailwind the distance required will be less, as a stronger tailwind is allowed for. If the wind is a 90 crosswind the distance required to take of will be the same as the distance calculated for zero wind component.RUNWAY SLOPEIf the runway is sloping a component of the weight will act along the longitudinal axis of the aeroplane. This will either augment thrust or augment drag which will increase or decrease the accelerating force. The amount of weight augmenting either thrust or drag is called either weight apparent thrust or weight apparent drag. It can be calculated by multiplying the force of weight by the sin of the angle of the runway slope.Figure 2.2 On a downslope a proportion of weight acts in the direction of thrustA downhill slope will increase the accelerating force, and reduce the take of distance, whereas an uphill slope will reduce the accelerating force and increase the take of distance.RUNWAY SURFACEEven on a smooth runway there will be rolling resistance due to the bearing friction and tyre distortion. If the runway is contaminated by snow, slush or standing water, there will be additional drag due to fuid resistance and impingement. This drag will increase with speed, until a critical speed is reached, the hydroplaning speed, above which the drag decreases. Despite this, during take-of, the hydroplaning speed is unlike to be reached, and as such any contamination will increase the drag and hence increase the take-of distance.146Chapter 2 General Principles -Take-OffIf the take-of is abandoned and braking is required, the coefcient of braking friction is severely reduced on a runway which is wet, icy or contaminated by snow or slush. This means that the brake pressure must be severely reduced to prevent skidding. Thus the accelerate stop distance is greatly increased.Figure 2.10.Figure 2.3 The efect of slush density on the slush dragAIRFRAME CONTAMINATIONThe performance data given assumes that the aircraft is not contaminated by frost, ice or snow during take of. In fact it is a requirement that at the commencement of take-of the aeroplane must be free of ice or snow. Snow and Ice on the airframe will increase the drag, reduced the lift and increase the weight of the aeroplane. Therefore, if any of these contaminants are present, the performance of the aircraft will be reduced, and the take-of distance will be increased.FLAP SETTINGFlaps afect: The CLmax of the wing The dragIncreasing fap angle increases CLmax, which reduces stalling speed and take of speed. This reduces the take of distance.Increasing fap angle increases drag, reducing acceleration, and increasing the take of distance. The net efect is that take of distance will decrease with increase of fap angle but above a certain fap angle the take of distance will increase again. An optimum seting can be determined for each type of aircraft, and any deviation from this seting will give an increase in the take ofdistance.Figure 2.11Figure 2.4 A graph showing the efect of the fap angle on the take-of distance required147Chapter 2 General Principles -Take-OffThe fap seting will also afect the climb gradient, and this will afect the Maximum Mass for Altitude and Temperature, which is determined by a climb gradient requirement, and the clearance of obstacles in the take-of fight path.Increasing the fap angle increases the drag, and so reduces the climb gradient for a given aircraft mass. The maximum permissible mass for the required gradient will therefore be reduced. In hot and high conditions this could make the Mass-Altitude-Temperature requirement more limiting than the feld length requirement if the fap seting for the shortest take-of distance is used. A greater take-of mass may be obtained in these conditions by using a lower fap angle.Figure 2.12 Optimum flap settingFigure 2.5 A graph showing the efect of fap on the climb and feld limit massIf there are obstacles to be considered in the take-of fight path, the fap seting that gives the shortest take-of distance may not give the maximum possible take-of mass if the Take-of Distance Available is greater than the Take-of Distance Required. If close-in obstacles are not cleared, using a lower fap angle will use a greater proportion of the Take-of Distance Available but may give a sufciently improved gradient to clear the obstaclesFigure 2.13 Optimum flap for obstacle clearanceFigure 2.6 An illustration showing the efect of fap angle on obstacle clearance148Chapter 2 General Principles -Take-Off149Chapter 2 General Principles -Take-OffQUESTIONS1. How is wind considered in the take-of performance data of the Aeroplane Operations Manuals?a. Unfactored headwind and tailwind components are used.b. Not more than 80% headwind and not less than 125% tailwind. c. Since take-ofs with tailwind are not permited, only headwinds are considered. d. Not more than 50% of a headwind and not less than 150% of the tailwind.2. What will be the infuence on the aeroplane performance if aerodrome pressure altitude is increased?

a. the total runway length, without clearway even if this one exists.b. the length of the take-of run available plus the length of the clearway available. c. the runway length minus stopway. d. the runway length plus half of the clearway.5. During take-of, the thrust of a fxed pitch propeller: a. increases slightly while the aeroplane speed builds up.b. varies with mass changes only. c. has no change during take-of and climb. d. decreases while the aeroplane speed builds up.6. What will be the efect on an aeroplanes performance if aerodrome pressure altitude is decreased? a. It will increase the take-of distance required.b. It will increase the take-of ground run. c. It will decrease the take-of distance required.d. It will increase the accelerate stop distance.150Chapter 2General Principles -Take-Off7. The take-of distance of an aircraft is 800m in a standard atmosphere with no wind and at 0 ft pressure altitude. Using the following corrections: 20 m / 1 000 ft feld elevation- 5 m / kt headwind+ 10 m / kt tail wind 15 m / % runway slope 5 m / C deviation from standard temperatureThe take-of distance from an airport at 2 000 ft elevation, temperature 21C, QNH 1013.25 hPa, 2% up-slope, 5 kt tail wind is:a. 810 mb. 970 m c. 890 m d. 870 m8. An uphill slope:a. increases the take-of distance more than the accelerate stop distance.b. decreases the accelerate stop distance only.c. decreases the take-of distance only.d. increases the allowed take-of mass.9. Other factors remaining constant and not limiting, how does increasing pressure altitude afect allowable takeof mass?a. Allowable take-of mass remains uninfuenced up to 5000 ft pressure altitude. b. Allowable take-of mass decreases. c. Allowable take-of mass increases. d. There is no efect on allowable take-of mass.10. In reality, the net thrust of a jet engine at constant RPM:a. does not change with changing altitude.b. is independent of the airspeed.c. decreases with the airspeed.d. increases with the airspeed.11. Which of the following are to be taken into account for the runway in use for take-of?a. Airport elevation, runway slope, standard temperature, pressure altitude and wind components.b. Airport elevation, runway slope, outside air temperature, standard pressure and wind components.c. Airport elevation, runway slope, outside air temperature, pressure altitude and wind components.d. Airport elevation, runway slope, standard temperature, standard pressure and wind components.12. For a take-of in slush, the slush drag:a. Will increase up to aquaplane speed and then remain constant.b. Will increase up to aquaplane speed and then decrease.c. Will increase up to aquaplane speed and then increase at a greater rate.d. Will decrease progressively up to the lift of speed.151Chapter 2 General Principles -Take-Off13. With contamination on the aircraft wings and fuselage only:a. The TODR will be unafected.b. The ASDR will decrease.c. Stalling speed is not afected.d. The lift of speed will be increased.14. The gross take-of distance (TOD) is defned as being from brake release until:a. The aeroplanes main wheel lifts of the runway.b. The aeroplane has reached 35 ft.c. The aeroplane has reached the screen height.d. The aeroplane is safely of the ground.15. The result of a higher fap seting up to the optimum at take-of is:a. a higher VR.b. a longer take-of run.c. a shorter ground roll.d. an increased acceleration.16. VR is the speed at which:a. The aeroplane nose wheel is of the ground.b. The pilot initiate the action required to raise the nose wheel of the ground.c. The main wheels lift of the ground.d. The aeroplane rotates about the longitudinal axis.17. During take-of:a. The acceleration force decreases.b. Wheel drag increases.c. Thrust increases.d. Total drag decreases.189. High altitudes, hot air and humid conditions will:a. Increase the take-of payload.b. Decrease the take-of mass.c. Increase the take-of performance.d. Decrease the take-of distance.19. The main purpose for taking of into wind is to:a. Decrease the true ground speed.b. Decrease the aeroplane performance.c. Increase the true ground speed.d. Increase the take-of distance.20. What is the efect of a contaminated runway on the take-of?a. Increases the take-of distance and greatly increases the accelerate stop distance.b. Increases the take-of distance and decreases the accelerate stop distance.c. Decreases the take-of distance and increases the accelerate stop distance.d. Decreases the take-of distance and greatly decreases the accelerate stop distance.152Chapter 2General Principles -Take-Off21. Which of the following statements is correct?a. If a clearway or a stopway is used, the lift-of point must be atainable at least by the end of the permanent runway surface.b. A stopway means an area beyond the take-of run available, able to support the aeroplane during an aborted take-of.c. An under-run is an area beyond the runway end which can be used for an aborted take-of.d. A clearway is an area beyond the runway which can be used for an aborted take-of.22. A Balanced Field Length is said to exist where:a. The accelerate stop distance available is equal to the take-of distance available.b. The clearway does not equal the stopway.c. The accelerate stop distance is equal to the all engine take-of distance.d. The one engine out take-of distance is equal to the all engine take-of distance.23. The stopway is an area which allows an increase only in:a. the accelerate-stop distance available.b. the take-of run available.c. the take-of distance available.d. the landing distance available.24. An airport has a 3,000 metres long runway, and a 2,000 meter clearway at each end of that runway. For the calculation of the maximum allowed take-of mass, the take-of distance available cannot be greater than:a. 4,000 metres.b. 3,000 metres.c. 5,000 metres.d. 4,500 metres.25. Can the length of a stopway be added to the runway length to determine the take-of distance available?a. Yes, but the stopway must be able to carry the weight of the aeroplane.b. Yes, but the stopway must have the same width as the runway.c. No.d. No, unless its centreline is on the extended centreline of the runway.26. In relation to runway strength, the ACN:a. Must not exceed 90% of the PCN and then only if special procedures are followedb. May exceed the PCN by up to 10% or 50% if special procedures are followedc. May exceed the PCN by a factor of 2d. Must equal the PCN27. The TODA is:a. Declared runway length onlyb. Declared runway length plus clearwayc. Declared runway length plus stopwayd. Declared runway length plus clearway and stopway153Chapter 2 General Principles -Take-Off28. Take-of distance available is?a. Take-of run available plus clearwayb. Take-of run minus the clearway, even if clearway existsc. Always 1.5 times the TORAd. 50% of the TORA.29. Can a clearway be used in the accelerate stop distance calculations?a. Yesb. Noc. Only if the clearway is shorter than the stopwayd. Only if there is no clearway30. The take-of distance available is?

a. The total runway length, without clearway even if this one exists. b. The length of the take-of run available plus any length of clearway available, up to a maximum of 50% of TORA. c. The runway length minus stopway. d. The runway length plus half of the clearway.31. The stopway is:a. at least as wide as the runway.b. no less than 152 wide.c. no less than 500 ft wide.d. as strong as the main runway .32. Which class of aeroplane describes all multi-engine turbo-jet aeroplanes?a. Unclassifed.b. Class C.c. Class B.d. Class A.33. A propeller aeroplane with nine or less passenger seats and with a maximum take-of mass of 5,700 kg or less is described as:a. Unclassifed.b. Class C.c. Class B.d. Class A.154Chapter 2General Principles -Take-OffANSWERS1 D 11 C 21 B 31 A2 A 12 B 22 A 32 D3 D 13 D 23 A 33 C4 B 14 C 24 D5 D 15 C 25 C6 C 16 B 26 B7 B 17 A 27 B8 A 18 B 28 A9 B 19 A 29 B10 C 20 A 30 B

Figure 3.4From Figure 3.4 it can be seen that the forward acting force in green, is now the same as the two rearward acting forces in red and the aeroplane will maintain a steady speed along its new fight path. To maintain a steady climb with no loss of speed, Thrust must balance not only the aerodynamic Drag, but also the backward component of Weight.160Chapter 3General PrinciplesClimb and DescentIn Figure 3.5 the aircraft only has a small amount of Excess Thrust available. Notice that there is too much backward component of Weight from the climb angle that has been set, so that climb angle can not be maintained.

THRUSTDRAGLIFTWEIGHTBACKWARD COMPONENTOF WEI GHT

Figure 3.5The angle of climb must be reduced to give a smaller backward component of Weight that matches the Excess Thrust available, as shown in Figure 3.6. The greater the Excess Thrust, the larger the backward component of Weight that can be balanced. THRUSTDRAGLIFTWEIGHTBACKWARD COMPONENTOF WEI GHTFigure 3.6In other words, the more Excess Thrust available, the steeper the angle of climb or the greater the weight at the same climb angle.161Chapter 3General PrinciplesClimb and DescentTHE EFFECT OF WEIGHT ON CLIMB ANGLETHRUSTDRAGWEIGHTI NCREASEDBACKWARD COMPONENT OF WEI GHTDUE TO HI GHER WEI GHTLIFTDECREASEDEXCESS THRUSTI NCREASEDTHRUSTBECAUSE OFI NCREASED DRAGTHRUSTDRAGWEIGHTLIFT Figure 3.7 Figure 3.8Weight has an infuence on climb performance. Figure 3.7 illustrates that if the aircraft tries to use the same climb angle as before, but at a higher weight, the backward component of weight will be greater and there is insufcient Excess Thrust to balance it. In addition, the higher weight will also generate increased aerodynamic Drag (Induced), which will further reduce Excess Thrust. Increased weight therefore decrease the maximum climb angle, as shown in Figure 3.8.THRUST AVAILABLETHRUSTDRAGLIFTEXCESS THRUSTTHRUST AVAI LABLEWEIGHTBACKWARD COMPONENTOF WEI GHTFigure 3.9Figure 3.9 shows that Thrust Available is the total amount of Thrust available from the engine(s) and under a given set of conditions and in a steady climb the thrust available must be the same as the sum of the aerodynamic Drag (D) plus the backward component of Weight (W sin ) .162Chapter 3General PrinciplesClimb and DescentCALCULATING CLIMB GRADIENTIf Thrust Available and aerodynamic Drag are known, the maximum backward component of Weight can be calculated by subtracting the aerodynamic Drag from Thrust Available. For most purposes, climb gradient is used rather than climb angle, but climb angle is still an important factor. Climb gradient is merely the percentage of the backward component of Weight to the aircraft Weight.Shown here is the formula to calculate the percentage climb gradient. This formula is very signifcant.

T - D Gradient % = x 100

WMerely by looking at the above formula certain facts are self evident: For a given weight, the greater the Excess Thrust (T D) the steeper the climb gradient. The less the Excess Thrust the more shallow the climb gradient. For a given Excess Thrust (T D), the greater the weight the more shallow the climb gradient. The less the weight the steeper the climb gradient.If representative values of Thrust, Drag and Weight are known the % climb gradient can be calculated. For example:A twin engine turbojet aircraft has engines of 60,000N each; its mass is 50 tonnes and it has a L/D of 12:1, what is the % climb gradient? Use g = 10m/s/s.From the above information the values to include in the formula have to be derived: Thrust = 60,000N x 2 engines = 120,000N Drag = Weight / 12 Weight = 50 tonnes x 1000 = 50,000kg x 10m/s/s = 500,000N Drag therefore = 500,000N / 12 = 41,667N

120,000N - 41,667 N

78,333 N

x 100 = x 100 = 15.7 %

500,000 N

500,000 NLet us now consider the same values, but with one engine failed: 60,000N - 41,667 N

x 100 = 3.7 %

500,000 NThrust has decreased by 50%, but climb gradient has decreased by approximately 75% or to one quarter of the gradient possible with all engines operating. This fact is very signifcant. After losing 50% of the Thrust Available, why did the gradient decrease by 75%?NOTE: The aircraft data gave the L/D ratio from which the value of Drag was extracted, yet the value of Lift is obviously less than Weight when the aircraft is in a steady climb. Isnt this an inaccurate value? If the climb angle is less than approximately 20 degrees, and it always will be, the diference in the magnitude of Lift and Weight in a steady climb is insignifcant and can be considered (For purposes of these and other calculations) to be the same.163Chapter 3General PrinciplesClimb and DescentCLIMBING AFTER AN ENGINE FAILUREConsider Figure 3.10: Losing 50% of the Thrust Available reduces Excess Thrust by approximately 75% because the same value of aerodynamic Drag must still be balanced. Figure 3.11 emphasises that a two engine aeroplane with one engine inoperative, has a severely reduced ability to climb.THRUSTDRAGWEIGHTONE ENGI NE I NOPERATI VEEXCESS THRUSTBOTH ENGI NETHRUST AVAI LABLELIFT

THRUSTDRAGWEIGHTLIFT

Figure 3.10 Figure 3.11164Chapter 3General PrinciplesClimb and DescentTHE EFFECT OF FLAPS ON CLIMBINGHigh Lift Devices (Flaps) increase aerodynamic Drag. From the study of Principles of Flight it was learned that the purpose of Flaps is to reduce the take-of and landing run. Figure 3.12 and Figure 3.13 show it is obvious that Flaps reduce the climb angle because they increase aerodynamic Drag and therefore decrease Excess Thrust.

DRAGTHRUSTLIFTFi gure 3. 12FLAPS REDUCEEXCESS THRUSTMORE DRAGFROM FLAPSTHRUST AVAI LABLEWEIGHTBACKWARD COMPONENTOF WEI GHT DRAGTHRUSTLIFTWEIGHTFLAPSREDUCECLI MB ANGLE Figure 3.12 Figure 3.13165Chapter 3General PrinciplesClimb and DescentTHE CLIMB ANGLE (FLIGHT PATH ANGLE) - GAMMAFigure 3.14 includes the symbol used for climb angle, the Greek leter GAMMA. Note that the angle between the horizontal and the fight path (Climb Angle) is exactly the same as the angle between the Weight vector and the transposed Lift vector. We will be using climb angle for the FREE AIR climb.THRUSTDRAGWEIGHTLIFTFigure 3.14When an aeroplane is in a steady climb there will be a gain in height after a given horizontal distance travelled. This relationship is the % Climb Gradient. The calculation on page 152 gave 15.7% climb gradient, all engines. From Figure 3.9 it can be visualised that for 100 units of horizontal travel, the aeroplane will be 15.7 units higher. This is a fundamental concept. For example: an aircraft with a climb gradient of 15.7% all engines operating, will be 314ft higher after travelling 2000ft horizontally, but the one engine inoperative climb gradient of 3.7%, will only give a height gain of 74ft in the same distance. Horizontal distance = 2000ft Gradient = 15.7% (For every 100ft horizontally, a height gain of 15.7ft) 200 ft = 20 100 20 x 15.7 = 314 ft (All engines) Gradient = 3.7% (For every 100ft horizontally, a height gain of 3.7ft) 20 x 3.7 = 74 ft (One engine inoperative)166Chapter 3General PrinciplesClimb and DescentPARASITE DRAG CURVEAn easy way to visualise how an aeroplane can maximize Excess Thrust and therefore its climb angle is to use a simple graph showing the relationship between Thrust and Drag under various conditions. But frst, the parts of the Drag curve will be studied in detail.PARASI TEDRAGI ASFigure 3.15Figure 3.15 shows Parasite Drag increasing with the square of the IAS. (Parasite Drag is proportional to IAS squared). If IAS is doubled, Parasite Drag will be increased four times. Note that at low speed Parasite Drag is small, but reaches a maximum at high IAS. Parasite Drag will increase with increasing Parasite Area (Flaps, undercarriage or speed brakes).INDUCED DRAG CURVEI NDUCEDDRAGI ASFigure 3.16Figure 3.16 shows Induced Drag decreasing with IAS squared. (Induced Drag is inversely proportional to IAS squared). If IAS is doubled, Induced Drag will be decreased to one quarter of its previous value). Note that Induced Drag is at its highest value at low IAS, but decreases with increasing IAS. Induced Drag will vary with Lift production. Increasing the Weight or banking the aircraft will increase Induced Drag.167Chapter 3General PrinciplesClimb and DescentTOTAL DRAG CURVEDRAGI ASorTHRUST REQUI REDTOTALDRAGPARASI TEI NDUCEDVmdMI NI MUMDRAGFigure 3.17In fight, an aeroplane will experience both Parasite and Induced Drag. The sum of Parasite Drag and Induced Drag is called Total Drag. When only the word Drag is mentioned, the meaning is Total Drag. Figure 3.17 shows a Total Drag curve: At any given IAS, Total Drag is the sum of Parasite Drag and Induced Drag. The IAS at which Parasite Drag is the same value as Induced Drag will generate minimum Total Drag. The IAS that gives minimum Total Drag, is called The Minimum Drag Speed and is called VMD. Flying at an IAS slower than VMD will generate more Drag and fying at an IAS faster than VMD will also generate more Drag.Whenever Drag is being considered it is an excellent idea to develop the habit of sketching the Total Drag curve, together with the Parasite Drag and Induced Drag curves. This enables the result of any variable to be included and the correct conclusion obtained.168Chapter 3General PrinciplesClimb and DescentTHE EFFECT OF WEIGHT OR BANK ANGLE ON DRAGI ASDRAGorTHRUST REQUI REDPARASI TEI NDUCEDHeavyTOTALDRAGHeavyLi ght VmdVmd HeavyMI NI MUMDRAGLi ghtMI NI MUMDRAGHeavyFigure 3.18Figure 3.18 shows the afect of increased Weight and/or the efect of turning the aircraft. Induced Drag will be greater at a given IAS because Lift must be increased when the aircraft has more weight or when turning. The reason for the proportionally greater increase in Induced Drag at the low speed end of the graph is because Induced Drag is inversely proportional to IAS squared therefore, at low speed the efect is greater.The intersection of the Induced Drag curve and the Parasite Drag curve is further towards the high speed end of the graph and the sum of Induced Drag and Parasite is greater. Total Drag will increase and VMD will be a faster IAS when an aircraft is operating at increased Weight or when turning,Conversely, throughout fight Weight will decrease due to fuel use. As the aircraft becomes lighter, Total Drag will decrease and the IAS for VMD will also decrease.169Chapter 3General PrinciplesClimb and DescentTHE EFFECT OF FLAPS OR GEAR ON TOTAL DRAG"Cl ean" VmdVmd - Gear & Fl apsMI NI MUMDRAG" Cl ean"MI NI MUMDRAGGear&Fl apsDRAGI ASorTHRUST REQUI REDGear&Fl apsTOTALDRAGPARASI TEGear&Fl apsI NDUCEDFigure 3.19Figure 3.19 shows the afect of Flaps or Gear (Undercarriage). Parasite Drag will be greater at a given IAS because the Parasite area will be increased. When both Flaps and the Gear are fully retracted, the aircraft is said to be in the Clean confguration or the aircraft is Clean. The reason for the proportionally greater increase in Parasite Drag at the high speed end of the graph is because Parasite Drag is proportional to IAS squared therefore, at high speed, the efect is greater. The intersection of the parasite Drag curve and the Induced Drag curve is further towards the low speed end of the graph and the sum of Parasite Drag and Induced Drag is greater. Total Drag will increase and VMD will be a lower IAS when either the Flaps or Gear are lowered. THRUSTThrust is the force required to balance aerodynamic Drag; plus the backward component of Weight when the aircraft is in a steady climb. A turbo-jet engine generates Thrust by accelerating a mass of air rearwards. The variation of Thrust Available with forward speed is relatively small and the engine output is nearly constant with changes in IAS.Thrust Available = Mass Flow x Acceleration (Exhaust velocity Intake velocity)Since an increase in speed will increase the magnitude of intake velocity, constant Thrust Available will only be obtained if there is an increase in Mass Flow or Exhaust velocity. When the aircraft is at low forward speed, any increase in speed will reduce the velocity change through the engine without a corresponding increase in Mass Flow and Thrust Available will decrease slightly. When the aircraft is fying at higher speed, the ram efect helps to increase mass fow with increasing forward speed and Thrust Available no longer decreases, but actually increases slightly with speed.170Chapter 3General PrinciplesClimb and DescentI ASTHRUSTAVAI LABLETURBO- JET THRUSTEssent i al l y const ant wi t h I ASFigure 3.20For a given engine RPM and operating altitude, the variation of turbo-jet Thrust with speed is shown in Figure 3.20.Because turbo-jet Thrust is essentially constant with speed, unless the take-of run is being considered, future illustrations will display Thrust Available from the turbo-jet as a straight line.VARIATION OF THRUST WITH DENSITY ALTITUDEA turbo-jet engine is un-supercharged. Increasing Density Altitude (lower air density) will reduce the mass fow through the engine and Thrust Available will decrease. Obviously, this will have an afect when climbing, but also when operating at airfelds with a High Pressure Altitude and/or a high Outside Air Temperature (OAT). As a reminder, Pressure Altitude can be determined on the ground by seting 1013mb on the altimeter sub-scale. If the altimeter reads 1000ft on the ground with 1013 on the sub-scale, the Pressure Altitude is 1000ft, irrespective of the actual height of the airfeld above sea level. (The aeroplane will experience the air pressure that corresponds to 1000ft in the International Standard Atmosphere).171Chapter 3General PrinciplesClimb and DescentLOW DENSI TY ALTI TUDEHI GH DENSI TY ALTI TUDEI ASTHRUSTAVAI LABLETURBO - JETFigure 3.21Figure 3.21 shows that Thrust Available has a lower value with increasing Density Altitude (Lower air Density).VARIATIONS OF TAKE-OFF THRUST WITH AIR TEMPERATURE (OAT)I SA +15 CTHRUSTAVAI LABLEThr ust i s " Fl at Rat ed" at l ower OAT Thr ust i s " EGT l i mi t ed" at Hi gher OAT GivenPressure

Altitude" Fl at Rat ed" THRUSTOutsi de Ai r Temperat ure ( OAT)C ( )"Ki nk"TURBO - JETFigure 3.22172Chapter 3General PrinciplesClimb and DescentGenerally, the Thrust of any turbo-jet engine is restricted by the maximum temperature the turbine blades can withstand. The more heat resistant the material from which the turbine blades are made and the more efcient the blade cooling, the higher the maximum turbine inlet temperature and therefore the greater the Thrust the engine can safely develop.For a given engine, the higher the OAT the lower the mass air fow and therefore the lower the fuel fow before the maximum turbine inlet temperature is reached and consequently, the lower the Thrust the engine is able to develop this is known as EGT limited Thrust.Figure 3.22 should be read from right to left, and shows Thrust increasing with decreasing OAT at a given Pressure Altitude, but only down to an OAT of ISA +15C. Below ISA + 15C Thrust remains constant. This is the engines Flat Rated Thrust. At OATs below ISA +15C, Thrust is no longer limited by turbine inlet temperature but by the maximum air pressure the compressor is built to withstand. Below airport OATs of ISA + 15C it does not mater how far the fight crew advance the throtle, the engine management computer will maintain Flat Rated Thrust this is the maximum certifed Thrust of the engine.From a Performance point of view, if engines are not Flat Rated and the throtles are fully advanced at OATs below ISA + 15C a lot more than maximum certifed Thrust will be delivered. While this may not be immediately destructive to the engine if done occasionally, it completely compromises the certifcation of the aeroplane. Engine-out critical speeds (VMCG, VMCA and VMCL) are based on the yawing moment generated at maximum certifed Thrust. If signifcantly more Thrust is produced during one-engine-out fight with the IAS at the recommended minimum, directional control of the aeroplane will be lost.Some Performance graphs incorporate the Flat Rated Thrust of the engine to allow determination of, for instance, the Climb Limit Take-of Weight. Climb Limit Take-of Weight will increase with decreasing OAT, but only down to ISA +15C. For each Pressure Altitude, an OAT lower than ISA +15C will not give an increase in Climb Limit Take-of Weight.In the JAA Performance exam the ISA +15C temperature for each pressure altitude is referred to as the kink in the pressure altitude lines of CAP 698, Figures 4.4, 4.5 and 4.29. The kink in the pressure altitude lines indicate the temperature, individually for each altitude, below which the Thrust will not increase with an increase in density.REGION OF REVERSE COMMANDTHRUST AVAI LABLE&THRUST REQUI REDI ASVmd350 KI ASTURBO - JETFigure 3.23173Chapter 3General PrinciplesClimb and DescentThrust and Drag on the same graph will show the result of many variables. Figure 3.23 shows Thrust Available in green and Thrust Required in red. The intersection of the two curves will result in unaccelerated fight at a high speed of 350 KIAS.THRUST AVAI LABLE&THRUST REQUI REDI ASVmd250 KI ASTURBO - JETFigure 3.24Figure 3.24 shows that to maintain unaccelerated fight at a lower speed of 250 KIAS, Thrust Available must be decreased and the aircraft slowed until Thrust Required reduces to the same value.THRUST AVAI LABLE&THRUST REQUI REDI ASVmdTURBO - JETFigure 3.25Figure 3.25 shows Thrust Available has been further reduced in order to maintain unaccelerated fight at Vmd, the minimum Drag speed.174Chapter 3General PrinciplesClimb and DescentTHRUST AVAI LABLE&THRUST REQUI REDI ASVmd175 KI ASTURBO - JETFigure 3.26Figure 3.26 shows that to maintain unaccelerated fight at an IAS slower than VMD, Thrust Available must be increased. This is because at speeds below VMD, Thrust Required (Drag) increases. The speed region slower than VMD has three alternative names: The back-side of the Drag curve, The speed unstable region and, perhaps the most descriptive, The region of Reverse Command: so called because to maintain unaccelerated fight at an IAS slower than VMD, Thrust must be increased the reverse of what is Normally required.175Chapter 3General PrinciplesClimb and DescentBEST ANGLE OF CLIMB SPEED (Vx)TOTAL DRAGorTHRUST REQUI REDVxVmd ( )MI NI MUMDRAGTHRUSTAVAI LABLE( Tur bo- Jet )MAXI MUMEXCESS THRUSTTURBO - JETTHRUST AVAI LABLE&THRUST REQUI REDI ASFigure 3.27Figure 3.27 shows Thrust Required (Aerodynamic Drag) and Thrust Available (From the engines) for an aeroplane powered by turbo-jet engines.Excess Thrust is the amount of Thrust that exceeds aerodynamic Drag. Excess Thrust can be seen on the graph as the distance between the Thrust Available and Thrust Required lines. You will recall that to maximise the climb gradient, Excess Thrust must be a maximum. Maximum Excess Thrust is obtained by fying at the IAS where the distance between the Thrust and the Drag lines is maximum.Notice that maximum Excess Thrust is available only at one particular IAS, labelled Vx. At any other speed, faster or slower, the distance between the Thrust and Drag curves is smaller and Excess Thrust is less. Therefore, climbing at an IAS other than Vx will give a climb gradient less than the maximum possible.The IAS at which the aeroplane generates the greatest amount of Excess Thrust and is therefore capable of its steepest climb gradient, is called Vx. (Vx is referred to as the Best Angle of Climb Speed). From Figure xx is can be seen that for an aeroplane powered by turbo-jet engines, Vx is the same IAS as VMD.176Chapter 3General PrinciplesClimb and DescentFACTORS AFFECTING ANGLE OF CLIMBTHE EFFECT OF WEIGHTViewing Thrust Required (Drag) and Thrust Available on the same graph will show any obvious changes in Excess Thrust and therefore maximum climb gradient. Any associated changes in the IAS for Vx can also be seen.THRUST AVAI LABLEANDTHRUST REQUI REDI ASTURBO - JETVx Vmd ( ) Li ghtVxVmd ( )HeavyMAXI MUMEXCESS THRUSTFigure 3.28Figure 3.28 shows the result of increased weight on the steady climb. More weight requires more Lift, therefore Induced Drag will be greater. This moves the Total Drag curve up, but also to the right. Thrust Required is increased and Vx is a faster IAS. Because Thrust Required has increased, Excess Thrust is decreased, so maximum climb gradient is decreased.Remember the formula to calculate climb gradient?

T D Gradient % = x 100

WMerely by looking at the above formula certain facts are self evident: For a given Weight, the greater the Excess Thrust (T D) the more times Weight will divide into the bigger value and therefore, the steeper the climb gradient. The less the Excess Thrust the more shallow the climb gradient. For a given Excess Thrust (T D), the greater the Weight the fewer times Weight will divide into the same value and therefore, the more shallow the climb gradient. The less the weight the steeper the climb gradient.Increased Weight reduces maximum climb gradient and increases Vx.177Chapter 3General PrinciplesClimb and DescentTHE EFFECT OF FLAPS (OR GEAR)Another factor that afects maximum climb angle is aircraft confguration. Confguration means whether the faps (or gear) are extended, or not. If neither faps (or gear) are extended the aircraft is said to be in the clean confguration. If faps (or gear) are extended Parasite Drag will increase, but there will be no signifcant change in Induced Drag.Vx - Fl aps & Gear Vmd ( )I ASTHRUST AVAI LABLEANDTHRUST REQUI REDTURBO - JETVx Vmd ( ) Cl eanMAXI MUMEXCESS THRUSTFigure 3.29Figure 3.29 shows a steady climb with faps (or gear) extended compared to the clean confguration. Parasite area is increased, therefore Parasite Drag will be greater. This moves the Total Drag curve up, but also to the left. Thrust Required is increased and Vx is a slower IAS. Because Thrust Required has increased, Excess Thrust is decreased, so maximum climb gradient is decreased.Flaps or gear reduce maximum climb gradient and decrease Vx.It therefore seems a very good idea to retract the gear as soon as possible after lift-of, after a positive rate of climb is achieved and also not to use faps during a climb so that the climb angle is as large as possible. But, you may recall the purpose of faps is to decrease the take-of and landing run. If it is necessary to use faps for the take-of run, retract them in stages after take-of as soon as it is safe to do so. The regulatory fap retraction schedule will be discussed later.178Chapter 3General PrinciplesClimb and DescentTHE AFFECT OF AIR DENSITY Air density afects the mass fow of air into the engine. A decrease in air density reduces Thrust Available, thus Excess Thrust is also decreased.Therefore, the ability to climb decreases with decreasing air density.Air density is presented on performance graphs as two components: Temperature and Pressure Altitude. (Pressure altitude is the reading on the altimeter when 1013mb is set on the sub-scale). Any variation in atmospheric pressure or temperature will change air density and therefore Excess Thrust.Relevant aircraft performance graphs contain a horizontal axis of temperature and a series of sloping Pressure Altitude guide lines. An intersection of these two values will provide the necessary compensation for Density Altitude.DENSITY ALTITUDENow might be a good time to review the meaning of Density Altitude. The Ofcial defnition can be confusing: A high density altitude is one that represents a higher altitude in the International Standard Atmosphere. A fair explanation when one already understands what is meant, but of litle instructional value. Air density cannot be Sensed or measured directly, but it can be calculated.Lets say that you are on an airfeld which is physically at sea level; the waves are lapping at the end of the runway. Due to existing meteorological conditions the air pressure is low (Lower than Standard sea level atmospheric pressure of 1013mb) and with 1013mb set on the altimeter sub-scale the altimeter reads 1000ft.The reason the altimeter reads 1000ft is because the actual air pressure is the same as that at 1000ft in the International Standard Atmosphere - you are at A high Pressure Altitude. Reduced air pressure will give reduced air density (Mass per unit volume). But air density is also afected by temperature. The Standard air temperature at 1000ft is 13 degrees C (Temperature lapse rate of 2 degrees C per 1000ft, so 15 2 = 13), yet for this example, the actual Outside Air Temperature is measured at 25 degrees C. The actual air temperature is 12 degrees C higher than Standard (25 13 = 12). This is referred to as ISA+12. Now, either a table or a circular slide rule can be used to accurately determine the Density Altitude; in this case: approximately 2400ft. So, although the aircraft is physically at sea level, the air density is the same as that at approximately 2400ft in the International Standard Atmosphere. A turbo-jet engine would theoretically generate less thrust and the TAS would need to be higher for a given IAS.For take-of and initial climb from the same airfeld, any increase in pressure altitude or air temperature due to local meteorological conditions will reduce Excess Thrust and therefore the ability to climb (or accelerate). This is in addition to the more obvious decrease in air density during a climb to high altitude.179Chapter 3General PrinciplesClimb and DescentVxVmd ( )THRUSTAVAI LABLE( Hi gh ai r densi t y)THRUSTAVAI LABLE( Low ai r densi t y)MAXI MUMEXCESS THRUSTTURBO - JETTHRUST AVAI LABLE&THRUST REQUI REDI ASFigure 3.30Any decrease in air density (Increase in Density Altitude) will reduce Thrust Available and therefore move the Thrust line downwards, as shown on the graph in Figure 3.30.Because decreased density reduces Excess Thrust, maximum climb angle will be reduced. Excess Thrust will continually decreases with increasing Density Altitude so the maximum angle of climb will continually decrease as the aircraft climbs. Note that Vx will remain constant with changes in air density, because at a constant IAS (Vx) Drag will not vary. However, you will recall that as air density decreases, True Air Speed must be increased to maintained the required dynamic pressure.So although the IAS for Vx is constant with increasing density altitude, the TAS for Vx will of course increase.You may recall from earlier lessons that high humidity will also decrease air density and will therefore also decrease aeroplane performance. This is already factored into performance charts, so is not something you need to allow for. However, basic theory questions in the exam may require your knowledge of the fact.THE AFFECT OF ACCELERATING ON CLIMBINGIt has been stated that the ability of an aircraft to climb depends upon Excess Thrust, which is the amount of Thrust Available remaining after Drag is balanced. Hence an aircrafts maximum climb angle is limited by its maximum Excess Thrust. If there is a need to accelerate the aircraft while climbing or a need to climb while accelerating, some of the Excess Thrust must be used for the acceleration and therefore the maximum climb angle will be reduced.180Chapter 3General PrinciplesClimb and DescentTHE AFFECT OF BANK ANGLE ON CLIMBINGWhen an aircraft is banked, any increase in bank angle beyond approximately 15 degrees will signifcantly increase the amount of Lift that needs to be generated. Increased Lift will generate more Induced Drag, so Excess Thrust will be reduced and therefore maximum climb angle will be reduced.THE AFFECT OF WIND ON CLIMBINGThe afect that wind has on climbing depends upon the type of climb gradient being considered, (wind being motion of a body of air over the ground). There are two types of climb gradient: Air gradient and Ground gradient. Air gradient is used by aviation authorities to lay down minimum climb performance limits. E.g. a Class A aeroplane: .. starting at the point at which the aeroplane reaches 400 ft (122 m) above the take-of surface, the available gradient of climb may not be less than 1.2% for two-engines aeroplanes.AIR GRADIENTAir gradient is the vertical distance gained in a body of air divided by the horizontal distance travelled through the same body of air. The fact that the body of air might be moving over the ground is NOT considered. So wind has no afect on Air gradient.BODY OF AI RFigure 3.31Figure 3.31 shows an aeroplane in the botom left corner of a body of air, directly above the control tower on the ground.BODY OF AI RSTI LL AI RaFigure 3.32Figure 3.32 shows the body of air stationary relative to the ground; this is referred to as Zero Wind or Still Air. The aeroplane has climbed to the top right corner of the body of air and the Air gradient is shown as Gamma a.Note: To simplify the study of climbing, for climb angles less than approximately 20 degrees, it is considered that doubling the climb angle will double the climb gradient.181Chapter 3General PrinciplesClimb and DescentGROUND CLIMB GRADIENTBODY OF AI RgggTAI LWI NDaI N A TAI LWI ND: <= AI R GRADI ENT ( Not af f ect ed by wi nd)= GROUND GRADI ENT ( I nf l uenced by wi nd)Al so known as t he Fl i ght Pat h Angl e ( FPA)Al so known as t he Cl i mb Angl e aaFigure 3.33Figure 3.33 shows the afect of a tailwind. Because the body of air is moving over the ground in the direction of fight the Ground gradient is smaller than the Air gradient.A tailwind does not change the Air gradient, but decreases the Ground gradient.BODY OF AI RgaI N A HEADWI ND: >a gHEADWI NDFigure 3.34Figure 3.34 shows the afect of a headwind. Because the body of air is moving over the ground opposite to the direction of fight the Ground gradient is larger than the Air gradient.A headwind does not afect the Air gradient, but increases the Ground gradient.The only time wind is used to calculate climb gradient is when obstacle clearance is being considered. In all other cases of climbing, still air is used, even if a wind value is supplied.182Chapter 3General PrinciplesClimb and DescentCALCULATING GROUND GRADIENTIt is possible to calculate the Ground gradient by using a wind factor to correct the Air gradient for wind. Any gradient is the vertical distance divided by the horizontal distance.BODY OF AI RgHEADWI NDa20kt1210020 80( TAS) 100( GS) 80= 1. 25 12% x 1. 25 = 15%Figure 3.35HEADWINDExample 1: An aeroplane has an Air gradient of 12%, its TAS is 100 kt and the headwind is 20 kt. Calculate the Ground gradient. (Figure 3.35).The 20 kt headwind makes the ground speed (GS) 80 kt (100 20 = 80).100 TAS divided by 80 GS gives a wind factor of 1.25. Multiplying the Air gradient of 12% by the wind factor gives a Ground gradient of 15%.Example 2: An Air gradient of 12% with a TAS of 160 kt and a headwind of 20 kt. 160 divided by 140 gives a wind factor of 1.14. Multiplying the Air gradient of 12% by the wind factor gives a Ground gradient of 13.7% (12 x 1.14 = 13.5).Note: It is important to remember that if the ground gradient is to be used for the calculation of obstacle clearance, the application of headwinds and tailwinds must include the 50% headwind and 150% tailwind rule.However, this will only be asked of you in the exam as a statement of fact and any climb questions of the above and similar variety will not require the 50% headwind and 150% tailwind rule to be applied. Obstacle climb questions requiring the use of the graphs in CAP 698 have the wind rules included in the graphs. E.g. CAP 698 Figs 4.20, 4.21.183Chapter 3General PrinciplesClimb and DescentTAILWINDBODY OF AI RgTAI LWI NDa20kt12( TAS) 100( GS) 120= 0. 83 12% x 0. 83 = 10%12020 100Figure 3.36Example 3: the same Air gradient of 12% with a TAS of 100 kt but now a tailwind of 20 kt. In the above example, the 20 kt tailwind makes the ground speed (GS) 120 kt(100 + 20 = 120).100 kt TAS divided by 120 kt GS gives a wind factor of 0.83. Multiplying the Air gradient of 12% by the wind factor gives a Ground gradient of 10% (12 x 0.83 = 9.96).Example 4: an Air gradient of 12% with a TAS of 160 kt and a tailwind of 20 kt makes the ground speed (GS) 180 kt. 160 kt TAS divided by 180 kt GS gives a wind factor of 0.89. Multiplying the Air gradient of 12% by the wind factor gives a Ground gradient of 10.7% (12 x 0.89 = 10.7).Having detailed the method of calculating Ground gradient from the Air gradient, we will now examine a typical climb gradient question.Example 5: Determine the ground distance for a Class B aeroplane to reach a height of 2000 ft above Reference Zero in the following conditions: OAT: 25C Pressure altitude: 1000 ft Gradient: 9.4% Speed: 100 KIAS Wind component: 15 kts HeadwindThe question mentions 2000 ft above Reference Zero; what is Reference Zero?Reference Zero is the point on the runway or clearway plane at the end of the Take-Of Distance Required (Figure 3.37). It is the reference point for locating the start point of the take-of Flight Path.BODY OF AI RTODRSCREEN REFERENCE ZEROFigure 3.37184Chapter 3General PrinciplesClimb and DescentBODY OF AI RHEADWI ND15kt10415 89( TAS) 104( GS) 89= 1. 17 9. 4% x 1. 17 = 11%SCREENFigure 3.38Figure 3.38 illustrates the data supplied in the question and calculation of the Ground Gradient from the Air gradient.1. The TAS is calculated from the KIAS using your circular slide rule. (at 1000ft Pressure altitude and 25 deg C, 100 KIAS = 104 KTAS)2. Due to the 15kt headwind, the Ground Speed will be (104 KTAS 15 Kt) = 89 KTAS.(Wind speed is always a TAS)3. TAS divided by GS gives a wind factor of 1.17.4. Multiplying the Air gradient by the wind factor gives a Ground gradient of 11% (Approximately).It is always a good idea to Draw the question. Once the triangle has been sketched and the known parameters included, the visual relationship can be more easily considered.BODY OF AI Rg1110050ftFigure 3.39185Chapter 3General PrinciplesClimb and DescentA Class B aeroplane is being considered, so the screen height is 50 ft. The climb segment begins at the screen height above Reference Zero. Therefore the aeroplane will only need to gain an additional 1950 ft to be 2000 ft above Reference Zero. An 11% gradient simply means that the aircraft will be 11 units higher after 100 units of horizontal travel.BODY OF AI Rg1950ft 11 = 177. 27100 x 177. 27 = 17727ft50ftFigure 3.40In this case the required vertical height gain is 1950 ft, so we need to discover how many times 11 will divide into1950 ft (1950 ft / 11 = 177.27). This means that the vertical height gain is 177.27 times greater, so the horizontal distance will also be 177.27 times greater. Multiplying 100 by 177.27 will give the horizontal distance travelled in feet, In this case, 17727 ft.Example 6: Following take-of, a light twin engine aeroplane has a 10% climb gradient. By how much will it will clear a 900m high obstacle situated 9740m from the end of the Take-ofDistance Available (TODA)?10 uni ts100 uni ts( 50ft ) 15m900mFigure 3.41In order to fnd out by how much the aeroplane will clear the obstacle, it is necessary to calculate the height gain after covering a horizontal distance of 9740m.Remember: Percentage gradient is merely the vertical height for a horizontal distance travelled of 100 units. In this case the 10% gradient will give 10 units up for every 100 units along.The primary data from the question has been included in Figure 3.41. This simple procedure allows you to see how the aircraft fight path relates to the obstacle. The height of the obstacle is above Reference Zero, as is the start of the climb segment.For practical purposes a screen height of 50ft is 15m.186Chapter 3General PrinciplesClimb and Descent10 x 97. 4 = 974m9740m 100 = 97. 4( 50ft ) 15m900mFigure 3.42The distance of the obstacle from the end of the TODA is 9740 m, so we need to discover how many times the horizontal ratio of 100 will divide into that distance (9740 / 100 = 97.4). This means that the horizontal distance is 97.4 times greater, so the height gain will also be 97.4 times greater. Multiplying 10 by 97.4 will give the height gain in metres, (97.4 x 10 = 974m) in this case, 974 m.However, it must be remembered that the climb segment starts at 15 m (50 ft) above Reference Zero. So the screen height must be added to the height gain (974 m + 15 m = 989 m), in this example, 989 mThe aircraft will clear the 900 metre obstacle by 89 metres.RATE OF CLIMBWe will now consider rate of climb and begin with an overview. There are many ways of learning. But once a concept has been explained and understood it must then be remembered. Some students manage to confuse angle of climb with rate of climb, so the following basic explanation is provided to help decide when and how to use rate of climb. (The same considerations can be used later with rate of descent).POWER is the RATE of doing work. (Associate the word RATE with the word POWER).Work = Force x Distance Therefore:

Force x DistancePOWER =

TimeWhen considering rate of climb we need to do the maximum amount of work on the aeroplane in a given time.Question: When climbing, what Force must be balanced?Answer: Drag!The remaining product from the formula is distance divided by time, e.g. nautical miles per hour (kt). Question: How many speeds are there?187Chapter 3General PrinciplesClimb and DescentAnswer: One! The True Air Speed, the only speed there is, the speed of the aeroplane through the air. Therefore: POWER REQUIRED = DRAG x TASIf we take a Thrust Required (Drag) curve in sea level ISA conditions, and multiply the Drag at various airspeeds by the TAS and plot the resulting Power Required curve on the same piece of graph paper, the result will be that illustrated in Figure 3.43.VmdThrustRequiredIASIASPowerRequiredVmpMI NI MUMPower Requi redFigure 3.43The shape of the Power Required curve is very similar to that of Thrust Required. The signifcant diference is that the Power Required curve is displaced to the left. Consequently, the speed for minimum Power Required (VMP) is slower than the speed for minimum Thrust Required (VMD).It is essential to be able to visualise the Power Required curve relative to the Thrust Required curve, together with the Vmp and Vmd relationship. Associated data will be presented later.To demonstrate one use of the (POWER REQUIRED = DRAG x TAS) formula; if an aircraft climbs at a constant IAS, Drag remains constant, but TAS must be increased to compensate for decreasing air density. Therefore, when climbing at a constant IAS, Power Required increases.Rate of climb is the vertical speed of an aeroplane measured in feet per minute; displayed in the cockpit on the vertical speed indicator (VSI). Another way to think of rate of climb is to consider it as the TAS of the aeroplane along a gradient. 188Chapter 3General PrinciplesClimb and DescentFigure 3.44 shows two identical aeroplanes at the same angle of climb. The one on the right has a higher TAS along the gradient. In the same time, the aeroplane on the right will climb through a greater vertical distance than the aeroplane on the left. Therefore the aeroplane on the right has a higher rate of climb. This demonstrates that TAS is one important factor when considering rate of climb.Figure 3.44Figure 3.45 shows two identical aeroplanes at the same TAS. The aeroplane on the right is climbing at a steeper angle. In the same time, the aeroplane on the right climbs through a greater vertical distance than the aeroplane on the left. Therefore the aeroplane on the right has a higher rate of climb. This demonstrates that angle of climb is also an important factor in the rate of climb.Figure 3.45From Figure 3.44 and 3.45 and the above explanation, it can be seen that rate of climb is a function of both angle of climb and TAS along the achieved gradient.189Chapter 3General PrinciplesClimb and DescentSAMPLE QUESTION An aircraft with a gradient of 3.3% is fying at an IAS of 85kt. At a Pressure Altitude of 8,500ft and an outside air temperature 15C, the aircraft will have a ROC of: a. 284 ft/min. b. 623 ft/min. c. 1,117 ft/min. d. 334 ft/min.As Power Required is Drag x TAS, the IAS must be converted into TAS at the pressure altitude of 8,500ft and an OAT of 15 degrees C. Using a circular slide rule the TAS is 100 KTAS.Note: At climb angles less than approximately 20 degrees (and they always will be) the diference in length between the hypotenuse and the adjacent sides of a right angled triangle is so small that, for the sake of simplicity, it is disregarded in this type of calculation. So we do not need to worry about the fact that the aeroplane TAS is up the hypotenuse. (The JAA make the same assumption, so your answers will be correct)From our previous study of climb angle/gradient it is self evident that use of percentage gradient allows us to visualise the ratio of up to along. In this case the climb gradient of 3.3% gives a horizontal component of 100 and a vertical component of 3.3. Because we are considering RATE of climb, the horizontal component is the TAS, which in this case is 100 KTAS; this must be converted into ft/min:- 100 KTAS x 6080 ft = 10133 ft/min

60 mins

10133 ft/min = 101.33

100

101.33 x 3.3 = 334 ft/minLet us return to the formula for the gradient of climb as shown below

(T D) Gradient =

WGradient is given by the formula thrust available minus thrust required divided by weight. All that it is needed now for consideration of Rate of Climb is to add the velocity function, as shown below. This is now the formula for Rate of Climb.

(T D) Rate of Climb = x TAS

W190Chapter 3General PrinciplesClimb and DescentHowever, there is a litle more detail to understand:-The velocity is True Air Speed, but Thrust and Drag are both forces and TAS is distance over time. Force multiplied by distance gives work and work divided by time gives power. This means that instead of thrust multiplied by velocity, the formula now contains the expression Power Available, and instead of thrust required multiplied by velocity, the formula now has Power Required.

(Excess power Available) Power Available - Power RequiredRate of Climb =

WThe Rate of Climb formula now reads power available minus power required (excess power), divided by weight. For any given weight, the greater the excess power available, the greater the rate of climb. Conversely, the less the excess power available, the smaller the rate of climb. In order to maximise the aeroplanes rate of climb therefore, we need to maximise excess power. To understand how it is possible to obtain the greatest amount of excess power available and therefore climb at the highest rate of climb it is necessary to look at some more graphs.191Chapter 3General PrinciplesClimb and DescentEXCESS POWER AVAILABLE (JET)Figure 3.46 shows a graph of power available and power required for a typical jet aeroplane. In order to provide some benchmarks it is necessary to locate the three reference speeds VMP, VMD

and 1.32 VMD that we mentioned earlier. The speed found at the botom of the power required curve is called the velocity for minimum power or VMP. There was another speed, slightly faster than VMP called VMD. This speed is the velocity for minimum drag and is found at the tangent to the power required curve. Lastly the highest of the performance reference speeds is 1.32 VMD which is located a point beyond VMD. Having analysed the reference speeds, the object now is to locate where excess power available is maximum. Figure 3.46 Maximum excess power available occurs at 1.32 VMD for jet aeroplanesLooking at the graph, the area between the two curves represents the area of excess power available. On the graph the greatest amount of excess power available will be found where the distance between the curves is at maximum. Notice that it occurs at 1.32 VMD. At any other speed, the excess power is less and the rate of climb will be less. The speed for the best rate of climb is called VY. Therefore for a jet aeroplane VY occurs at 1.32 VMD. VY is the airspeed to use to climb to the cruise or en-route altitude as it will give the greatest height gain per unit time. In a typical 737-400 this speed is about 275 knots and is usually published in the aeroplane fight manual as an indicated airspeed.192Chapter 3General PrinciplesClimb and DescentEXCESS POWER AVAILABLE (PROPELLER)Figure 3.47 is a graph of power available and power required for a typical propeller aeroplane. On the graph, the greatest amount of excess power available will be found where the distance between the curves is at maximum. Figure 3.47 Maximum excess power available occurs at VMD for propeller aeroplanesNotice that for a propeller aeroplane this occurs at VMD. At any other speed, the excess power is less and the rate of climb less. As we have already learnt, VY is the speed for the best rate of climb. Therefore for a propeller aeroplane, VY occurs at VMD.It is important to remember where VY is found for both jet and propeller aeroplanes. For example, looking at the Figure 3.47, it can be seen that if a propeller aeroplane were climbing at a speed equal to VMP and then selected a slightly higher speed, the excess power would increase and the rate of climb would increase. It is more obvious to see how speed changes afect both excess power and the rate of climb when the graph is used. Try and become aware of what the graphs look like so that you can use them to your advantage in test situations. Let us now examine what factors can infuence the rate of climb for jet or propeller aeroplanes.FACTORS AFFECTING RATE OF CLIMBWEIGHTYou may recall that an increase in weight creates more weight apparent drag which reduces the angle of climb. For any given airspeed, if the angle of climb reduces, then so will the rate of climb because they are fundamentally linked. You can see this efect on the formula shown below. Simply by increasing the value of weight, mathematically the rate of climb will reduce.

Power Available - Power RequiredRate of Climb =

W193Chapter 3General PrinciplesClimb and DescentWeight has a further efect that we have already talked about. An increase in weight will require an increase in lift. Increasing lift increases induced drag which causes the drag curve to move up and right. The power required curve, shown in red in the graph below, is actually based upon drag. Power required is drag multiplied by velocity.So if the drag curve moves up and right, so will the power required curve. Figure 3.48 The efect of higher weight is to move the power required curve up and right, reducing excess thrust and increasing VYFigure 3.49 The efect of higher weight is to move the power required curve up and right, reducing excess thrust and increasing VY194Chapter 3General PrinciplesClimb and DescentAs you can see from Figure 3.48 for jet aeroplane and Figure 3.49 for propeller aeroplanes the power required curve moves up and right. Therefore less excess power is available and therefore the rate of climb decreases. However, what is important to see is that the speed for maximum excess power available is no longer the same. It is now higher. So with higher weight, the rate of climb is decreased but VY is increased.CONFIGURATION The next factor that afects the rate of climb is the confguration of the aircraft, in other words the use faps and undercarriage. If the undercarriage and faps are deployed then the profle drag of the aeroplane will increase. This increases total drag and the drag curve moves upwards and to the left. The power required curves will follow the same movement as the drag curve. Figure 3.50 The efect of extending the gear or faps is to move the power requiredcurve up and left, reducing excess thrust and decreasing VYNotice the reduction in excess power available as shown by the blue double headed arrows. You may recall that a reduction in excess power reduces your rate of climb. However, the important efect here is that the speed for maximum excess power is no longer the same. It is now lower. So with the undercarriage and or faps deployed, the rate of climb is decreased and VY is decreased. If you use faps for take of, remove them in stages when you have atained a positive stable climb ensuring you check through the aeroplane fight manual for the correct actions for your aeroplane.As you do so, the rate of climb and the speed to atain the best rate of climb will increase, so you should accelerate to ensure you remain at VY.DENSITYDensity is another important factor that afects the rate of climb. However, density afects a lot of the variables in the formula for the rate of climb.Shown below is an expanded rate of climb formula, reminding you that power available is thrust multiplied by true airspeed and that power required is Drag multiplied by true airspeed.

W195Chapter 3General PrinciplesClimb and DescentFocusing on the power available for the moment, decreased density will decrease the thrust but it will also increase the true airspeed. The overall efect is that the thrust loss is more than the TAS gain, meaning, overall, that the power available decreases.Looking at power required, decreased density will increase the true airspeed but have no efect on the drag. Therefore the power required will increase. Looking at Figure 3.51and using the graphs for the Jet and Propeller aeroplane you can see that the power available curve move down and right, and the power required curve move up and right. You will notice that there is less excess power available and this causes a reduction in the rate of climb for both aeroplane types. Figure 3.51 Decreasing density (high temperature/high altitudes/high humidity) reduces excess power available and increases VY (TAS)Notice from the graph that the true airspeed for VY increases a litle with decreasing density or increasing altitude. However, as pilots we fy using indicated airspeeds and therefore it is important to understand what happens to the indicated airspeed of VY. In order that we may understand a further explanation is needed.Using Figure 3.52 youll notice that if the true airspeed increases only slightly with altitude, then the indicated airspeed will still fall. Therefore, although VY as TAS increases with decreasing density or increasing altitude, VY as an IAS decreases. In fact, VY will eventually fall to become the same value as VX. So in summary, reduced density decreases the indicated airspeed of VY

and decreases the rate of climb.

Figure 3.52 If the TAS of VY increases a litle with reducing density or increasing altitude the IAS of VY still falls196Chapter 3General PrinciplesClimb and DescentIn relation to altitude, the higher the aeroplane fies, the excess power available diminishes and therefore the maximum achievable rate of climb will decrease.There will be an altitude where the excess power available decreases to zero, as shown in Figure 3.53. Therefore the rate of climb will also decrease to zero. This altitude is known as the absolute ceiling. Figure 3.53 At a certain altitude or density, there is no more excess power available and therefore the rate of climb is zeroFigure 3.52 shows the excess of power for a typical aeroplane at various altitudes. Notice VY

is the speed that gives the maximum excess power available and maximum achievable rate of climb. This is shown by the top of each curve. Also note that on this graph, VX can be found at the tangent point of each curve. As altitude increases, notice that the excess power available, achievable rate of climb and the indicated airspeed for VY decreases. Eventually there will be an altitude where VX and VY are the same and there is no more excess power and therefore, the rate of climb is zero. Remember that this altitude is called the Absolute Ceiling. Figure 3.54 As altitude increases, the excess thrust reduces and VY as an IAS decreases to become the same speed as VX at the absolute ceilingAt its absolute ceiling, the performance of an aeroplane is so reduced that it is unable to manoeuvre. Therefore, absolute ceiling is a rather abstract concept for a pilot. It is more useful for a pilot to know his aeroplanes service ceiling. Service ceiling is defned by the manufactures and aviation authorities as the maximum altitude where the best rate of climb airspeed will still produce a positive rate of climb at a specifc number of feet per minute. The recommendation is to not exceed this altitude because the performance envelope of the aeroplane is very small. If the aeroplane were to climb higher, the rate of climb would fall to zero, the aeroplane will not be able to climb any higher and the absolute ceiling would be reached. At this altitude VY and 197Chapter 3General PrinciplesClimb and DescentVX are the same speed. To fnd the absolute and service ceilings of your aeroplane, consult you fight manual and operate the aeroplane below the service ceiling to help maintain sufcient performance levels. WINDWind is a generally considered as horizontal movement of air. It cannot oppose or add to the vertical forces on the aeroplane. As such, wind has no efect on the rate of climb and therefore has no efect on the time to climb. However, vertical wind currents such as the ones found in microbursts or vertical windshear do afect the rate of climb of the aeroplane. DESCENTThis lesson on descent performance will focus mainly on the forces in the descent and what factors govern the descent. In a normal fight, the descent will occur at a point we defne as the top of descent which may be up to 200 miles before the destination aerodrome. A descent will also be required following engine failure or depressurisation. In this later situation the descent is forced early and it is important for the pilot to be aware of what determines the characteristics of the descent so that obstacle clearance can be maintained. There are two ways of measuring the descent performance of an aircraft. Either by angle of descent, sometimes called descent range, or rate of descent, sometimes called descent endurance.ANGLE OF DESCENTIn order to initiate a steady descent, thrust is normally reduced. The rearward force of drag exceeds the forward force of thrust and the aircraft slows down. The value of drag that exceeds the thrust force is called excess drag. In order to balance the forces and maintain speed, the nose is lowered until the weight apparent thrust provides enough forward force to balance the excess of drag as can be seen in Figure 3.55. Now the aircraft will maintain this steady descent angle at a constant speed. The forward and rearward forces are in balance once again. Drag (DA) is being balanced by the thrust (T) and the weight apparent thrust (W sin ). DA = T + W sin Figure 3.55 An illustration showing the balances of forces in normal powered descent198Chapter 3General PrinciplesClimb and DescentThe weight apparent thrust can be calculated by multiplying weight by the sine of the angle gamma. If thrust were reduced even more as shown in Figure 3.56, then there would be a greater amount of excess drag. More weight apparent thrust is therefore needed to balance the greater amount of excess drag. To gain more weight apparent thrust the aeroplane nose must be lowered even more. The result is an increase in the descent angle. For the purpose of the examinations, lowering the nose is a decrease in pitch.Figure 3.56 An illustration showing the balances of forces in low powered descent.From this demonstration it is clear that it is the excess drag which determines the angle of descent. Notice that the angle gamma is the same angle as the angle of descent. Re-arranging the formula shown in Figure 3.55 and 3.56 (DA = T + W sin ) so that angle gamma can be calculated gives us the formula for the angle or gradient of descent.

D - TGradient of Descent(%) = x 100

WDrag minus Thrust will give Excess drag. In summary then, the angle or gradient of descent is controlled by the excess drag. In order to visualise this excess drag, it is necessary to return to the thrust and drag graphs that were used in the climbing lesson.199Chapter 3General PrinciplesClimb and DescentFigure 3.57 An illustration showing excess drag (D T) for a jet and propeller aeroplaneShown in Figure 3.57 are the thrust and drag curves for a jet and propeller aeroplane. We have learnt that in order to descend, there has to be an excess of drag. On the graphs, excess drag can be found by taking the area beneath the drag curve and subtracting from it the area beneath the thrust curve. The solid blue highlighted areas represent excess drag. Notice that if thrust is reduced at any given speed, then excess drag increases, therefore the descent angle increases. MAXIMUM ANGLE OF DESCENTIf it were a performance priority to maximise the angle of descent, then from the theory we have seen so far, we would have to maximise excess drag. Figure 3.58 An illustration showing excess drag (D T) for a jet and propeller aeroplane with zero thrustIf thrust is reduced to zero, notice that the excess drag area is a maximum, but to obtain the maximum excess drag, the aeroplane needs to be accelerated to a very high speed, as shown in Figure 3.56. This can be achieved by closing the throtles, and continuously lowering the nose of the aeroplane so as to cause the increasing amount of weight apparent thrust to accelerate the aeroplane. As speed rises even more, both the excess drag and the angle of descent will increase. The angle of descent then is a function of excess drag, the greater the excess drag the steeper the angle of descent. This angle can be increased even more if it were possible to increase the excess drag. This can be achieved by deploying drag devices such as the speed brakes and the undercarriage, but atention must be paid to their maximum deployment speeds. The practical side of increasing drag to increase your descent angle occurs mainly in training and occasionally in commercial operations after air trafc re-routes. Take note that any increase in excess drag either by deploying faps or undercarriage, or by reducing thrust will steepen the angle of descent.200Chapter 3General PrinciplesClimb and DescentMINIMUM ANGLE OF DESCENTYou have just learnt that to maximise the descent angle excess drag must be maximised. Following an engine failure the aim is to ensure that the aeroplane will cover the greatest horizontal distance so that the pilot has a large area in which to select a suitable landing feld. This can only be achieved by using the fatest possible angle of descent sometimes referred to as the minimum glide angle. To descent at the shallowest possible angle, then excess drag must be minimum. This can be seen in Figure 3.59.Figure 3.59 An illustration showing that excess drag (D T) is minimum at VMD for a jet and propeller aeroplane with zero thrustNotice that VMD is the speed found at the botom of the drag curve. At VMD, drag is at a minimum. Therefore, to descent at the minimum possible angle, the aeroplane must be fown at VMD since, with no power, VMD is the speed that gives minimum excess drag.One of the other ways to examine glide performance is to consider the aeroplanes lift over drag ratio commonly referred to as the lift drag ratio. However, before we do this, there is a litle more detail to discuss frst. In the glide descent the resultant of drag and lift balances the force of weight. From your Principles of Flight lessons, the term for the resultant of lift and drag is Total Reaction. If the drag force line is moved upwards you can see that we now have a triangle of forces with the angle gamma being the angle between the lift and total reaction.Figure 3.60 The value of drag (in red) and the value of lift (in yellow) determine the angle gamma, which is also the same as the angle as the angle of descent.201Chapter 3General PrinciplesClimb and DescentLooking at Figure 3.60 if there were any change in the lift and or drag values, both the angle gamma and the glide angle would change. A typical modern jet has a maximum lift drag ratio of about 19. Where this ratio reaches its maximum, the value the angle gamma will be at minimum. The lift drag ratio reaches its maximum value at 4 degrees angle of atack and is always at the speed VMD. Therefore, VMD is the speed for the minimum angle of descent or the minimum glide angle. VMD is also known as the speed for L/D Max or L/D Min. The angle of atack at VMD is fxed at 4 degrees. If the angle of atack is greater or less than 4 degrees, the speed will change, the lift drag ratio will decrease in value and consequently the angle of the glide will increase.If, following an engine failure, you fy at VMD, your aeroplane will be fying at the fatest possible glide angle. Never try to stretch the glide by raising the nose of the aeroplane. If you do so, the speed will decrease and the glide angle will steepen. At any speed other than VMD your glide angle will be steeper than the optimal glide angle.Shown in Figure 3.61 is the angle of glide at VMD.Figure 3.61When fying at VMD, is the nose is raised the aeroplane will slow down and therefore not be at VMD anymore. The angle of descent will therefore steepen.The temptation is to raise the nose a litle. This gives an impression from the cockpit that the glide is being extended. If the nose were raised, the weight apparent thrust would decrease and the aeroplane would slow down. As a result the aeroplane would not be fying at VMD, and therefore not have the best lift drag ratio. The result would be a steeper descent angle despite the fact that the aeroplane had a slightly higher nose atitude. Be disciplined in following the procedures for fying at the optimum glide angle as laid down by the manufactures or the operator in the Aeroplane Flight Manual.202Chapter 3General PrinciplesClimb and DescentRATE OF DESCENTYou may recall we mentioned at the beginning of the descent section that there were two ways of assessing the descent performance. We have covered the angle of descent or descent range; now let us consider the rate of descent or descent endurance.You have already learnt in an earlier lesson that the rate of climb was a function of both climb angle and velocity. Similarly, the rate of descent is a function of descent angle and velocity. Shown below is the formula for the rate of descent.

DV - TVRate of Descent =

WYou may recall that force and velocity gave us Power. So the correct formula for the rate of descent is the shown below.

(Excess power Required) Power Required - Power AvailableRate of Descent =

W

The rate of descent is equal to the power required minus the power available divided by the weight. Power required minus power available gives excess power required. So, for any given weight, the rate of descent is determined by the excess power required. The greater the excess power required, the larger the achievable rate of descent, conversely the lesser the excess power required the smaller will be the rate of descent.In an emergency descent, for instance a descent initiated by the pilot following depressurisation, the aim is to reach FL 100 as soon as possible. To achieve this requirement, the aeroplane would need to lose height with the maximum possible rate of descent which can only happen if the aeroplane has maximum excess power required. Shown in Figure 3.62 is the power required and power available curves for both the jet and propeller aeroplane. The areas beneath the power required curves but above the power available curves represent the excess power required and are highlighted in light blue. However, notice that is not a lot of excess power required. It is possible to create even more.Figure 3.62 The area beneath the power required line but above the power available line represents the area for excess power required. This is shown by the blue faded areas.203Chapter 3General PrinciplesClimb and DescentIn order to create the conditions of the greatest excess of power required, power available (represented by the green line in the diagram) must be reduced to nil as shown by Figure 3.61 Figure 3.63 With no power available since the throtle is closed, all of the area beneath the power required curve becomes excess power requiredNotice that now, the blue areas of excess power required, are now as large as possible. In order to achieve the greatest rate of descent the aeroplane needs to fy at a speed that achieves the greatest excess power required. It is obvious from the graph that this can only be achieved at very high speeds. Remember that power required is a function of speed and drag, therefore the aeroplane needs to be confgured for high drag and high speed in order to achieve maximum power required. In fact, for a lot of Class A aeroplanes, emergency descents are fown at maximum operating speeds with air brakes or speed brakes deployed and thrust at idle.If it were the aim to descend at the lowest rate of descent, the aeroplane would need to fy at a speed that gives the minimum excess power required. Looking at Figure 3.63 this point is plain to see. In fact it is found at the very botom of the power required curve. You may recall that this speed is called VMP. So, to lose height at the slowest possible rate of descent the aeroplane would need to fy at VMP. The lowest rate of descent is also known as maximum descent endurance, which essentially means the aeroplane will take the greatest time to descend. The JAA sometimes refer to this as the speed for maximum glide endurance.204Chapter 3General PrinciplesClimb and DescentFACTORS AFFECTING DESCENTWEIGHTFor the efect of weight on the descent we shall only consider the efect in a glide, in other words with idle power. Let us frstly concentrate on the minimum angle of descent or the glide angle.Figure 3.64 An illustration showing that with a higher weight the forward and rearward forces along the fight path are larger but the angle of glide remains unchangedLooking at Figure 3.64 you can see that an aeroplane with a higher weight will have a larger amount of weight apparent thrust, but if the aeroplane is still fying at VMD, (which will be faster with a higher weight) it will also have a greater amount of drag. You will recall this from knowing that a higher weight moves the drag curve up and right. In Figure 3.64 youll notice that the forward and rearward forces along the fight path are still balanced, albeit a bit longer. But crucially notice that the angle of descent is unchanged. It is important for you to understand that weight has no efect on the minimum angle of descent or glide angle but it will increase the speed of the descent. In summary therefore, weight has no efect on the minimum angle of descent, but it will increase the speed along that descent gradient and therefore it will increase the rate of descent.205Chapter 3General PrinciplesClimb and DescentCONFIGURATIONSimilarly with the efect of weight, confguration changes are best understood when assuming idle thrust. The next factor to afect the angle and rate of descent is the aeroplanes confguration. Looking at Figure 3.65, if the faps or undercarriage were deployed, then notice that excess drag increases. To balance this increase in excess drag, the nose is lowered. This action increases weight apparent thrust and a balance of forces is restored but, importantly the balance is achieved at a higher angle of descent and therefore a higher rate of decent.Figure 3.65 An illustration showing an increase in excess drag must be balanced by an increase in weight apparent thrust.The efect of confguration can also be seen using graphs. Shown below in Figure 3.66 is the drag curve for the jet and propeller aeroplane with excess drag shown by the light blue area. With faps and undercarriage deployed, you will recall that the curves move up and left. This has the efect of increasing the excess drag and therefore increasing the angle of descent for any given speed. Notice too, that the speed for the minimum angle of descent, VMD, is lower.Figure 3.66 With undercarriage and fap deployed, the drag curve moves up and left showing an increase in excess drag and consequently an increase in glide angle.206Chapter 3General PrinciplesClimb and DescentThe same efect can be seen in Figure 3.67 when examining the rate of descent and by using the power required graph. Similarly the blue area represents excess power required. Figure 3.67 With undercarriage and fap deployed, the power required curve moves up and left showing an increase in excess power required and consequently an increase in the rate of descent.With faps and undercarriage deployed, you will recall that the power curves move up and left. This has the efect of increasing the excess power required and therefore increasing the rate of descent. Notice too, that the speed for the minimum rate of descent, VMP, is lower.In summary then, with gear and faps deployed the angle and rate of descent increase, but the speeds for minimum angle and minimum rate of descent decrease.207Chapter 3General PrinciplesClimb and DescentWINDThe last factor to afect the angle and rate of descent is wind. Figure 3.68 shows the efect of headwinds and tailwinds on the angle and rate of descent. Headwinds steepen the glide angle and decrease the descent range whereas tailwinds decrease the glide angle but increase the descent range. However, notice that the aeroplane in a headwind or tailwind reaches the same descent altitude in the same time as the aeroplane fying in zero wind conditions. This demonstrates that a headwind or tailwind has no efect on the rate of descent. Figure 3.68 An illustration showing the efect of headwinds and tailwinds on the angle and rate of descent.The efect of the wind on the angle of descent can be examined a litle further. Because of the adverse efect of the headwind on descent range, then in a glide, it would be of a beneft to increase the aeroplanes forward speed slightly. This has the efect or reducing the time spent in the head wind. This means that the aeroplane will not be pushed back as much by the wind. Similarly with a tailwind; because a tailwind benefts the glide by increasing the decent range, it would be beter to try and stay in this situation for longer. So this time the aeroplanes forward speed can be decreased so that the aeroplane can stay under the tailwind efect for longer and therefore be pushed further forwards.When fying on a training sortie make sure you know the wind speed and direction both for the surface and aloft. This will help you plan a beter descent giving you more accurate circuit paterns. But more importantly, knowledge of what the wind is doing will ensure you obtain maximum descent performance if an engine failure should occur.208Chapter 3General PrinciplesClimb and DescentCLIMB QUESTIONS1. What happens to the drag of a jet aeroplane if, during the initial climb after take of, a constant IAS and constant confguration is maintained? (Assume a constant mass..a. The drag decreases.b. The drag increases initially and decreases thereafter.c. The drag remains almost constant.d. The drag increases considerably.2. The speed for best rate of climb is called?a. VO.b. VY.c. VX.d. V2.3. An increase in atmospheric pressure has, among other things, the following consequences on take-of performance:

a. a reduced take-of distance and degraded initial climb performance.b. a reduced take-of distance and improved initial climb performance.c. an increases take-of distance and degraded initial climb performance.d. an increased take-of distance and improved initial climb performance.4. A higher outside air temperature:a. does not have any noticeable efect on climb performance.b. reduces the angle of climb but increases the rate of climb.c. reduces the angle and the rate of climb.d. increases the angle of climb but decreases the rate of climb.5. In un-accelerated climb:a. thrust equals drag plus the uphill component of the gross weight in the fight path direction.b. thrust equals drag plus the downhill component of the gross weight in the fight path direction.c. lift is greater than the gross weight.d. lift equals weight plus the vertical component of the drag.6. A jet aeroplane is climbing at a constant IAS with maximum climb thrust.How will the climb angle / the pitch angle change?a. Remain constant / decrease.b. Remain constant / become larger.c. Reduce / decrease.d. Reduce / remain constant.209Chapter 3General PrinciplesClimb and Descent7. Take-of performance data, for the ambient conditions, show the following limitations with fap 10 selected:Runway or Field limit mass: 5,270 kgObstacle limit mass: 4,630 kgIf the estimated take-of mass is 5,000kg it would be prudent to consider a take-of with faps at:a. 20, both limitations are increased.b. 5, the obstacle limit mass is increased but the runway limit mass decreases.c. 5, both limitations are increased.d. 20, the obstacle limit mass is increased but the runway limit mass decreases.8. A four jet-engined aeroplane whose mass is 150 000 kg is established on climb with engines operating. The lift over drag ratio is 14:1. Each engine has a thrust of 75 000 Newtons. The gradient of climb is: (given: g = 10 m/s.a. 12.86%b. 27%c. 7.86%d. 92%9. How does the best angle of climb and best rate of climb vary with increasing altitude?a. Both decrease.b. Both increase.c. Best angle of climb increases while best rate of climb decreases.d. Best angle of climb decreases while best rate of climb increases.10. Following a take-of determined by the 50 ft (15m) screen height, a light twin climbs on a 10% ground gradient. It will clear a 900 m high obstacle situated at 10,000 m from the 50 ft clearing point with an obstacle clearance of:a. 85 mb. It will not clear the obstaclec. 115 md. 100 m11. The rate of climb:a. is approximately the climb gradient multiplied by the true airspeed divided by 100.b. is the downhill component of the true airspeed.c. is angle of climb multiplied by the true airspeed.d. is the horizontal component of the true airspeed.12. Assuming that the required lift exists, which forces determine an aeroplanes angle of climb?a. Thrust and drag only.b. Weight and thrust only.c. Weight, drag and thrust.d. Weight and drag only.210Chapter 3General PrinciplesClimb and Descent13. Which of the following provides maximum obstacle clearance during climb?a. 1.2Vs.b. The speed for maximum rate of climb.c. The speed, at which the faps may be selected one position further UP.d. The speed for maximum climb angle Vx.14. Which speed provides maximum obstacle clearance during climb?a. The speed for which the ratio between rate of climb and forward speed is maximum.b. V2 + 10 kt.c. The speed for maximum rate of climb.d. V2.15. Which of the equations below expresses approximately the unaccelerated percentage climb gradient for small climb angles?a. Climb Gradient = ((Thrust - Drag./Weight. x 100b. Climb Gradient = ((Thrust + Drag./Lift. x 100c. Climb Gradient = ((Thrust - Mass./Lift. x 100d. Climb Gradient = (Lift/Weight. x 10016. The absolute ceiling:a. is the altitude at which the best climb gradient atainable is 5%.b. is the altitude at which the aeroplane reaches a maximum rate of climb of 100 ft/min.c. is the altitude at which the rate of climb is theoretically zero.d. can be reached only with minimum steady fight speed.17. The climb gradient of an aircraft after take-of is 6% in standard atmosphere, no wind, at 0 ft pressure altitude.Using the following corrections: 0.2 % / 1 000 ft feld elevation 0.1 % / C from standard temperature- 1 % with wing anti-ice- 0.5% with engine anti-iceThe climb gradient after take-of from an airport situated at 1 000 ft, 17 C; QNH 1013.25 hPa, with wing and engine anti-ice operating for a functional check is:a. 3.9 %b. 4.3 %c. 4.7 %d. 4.9 %18. As long as an aeroplane is in a positive climb:a. VX is always below VY.b. VX is sometimes below and sometimes above VY depending on altitude.c. VX is always above VY.d. VY is always above VMO.211Chapter 3General PrinciplesClimb and Descent19. A constant headwind component:a. increases the angle of fight path during climb.b. increases the best rate of climb.c. decreases the angle of climb.d. increases the maximum endurance.20. A higher gross mass at the same altitude will cause:a. VY and VX to decrease.b. VX to increase and VY to decrease.c. VY and VX to remain constant since they are not afected by a higher gross mass.d. VY and VX to increase.21. With an true airspeed of 194 kt and a vertical speed of 1,000 ft/min, the climb gradient is approximately:a. 3b. 3%c. 5d. 8%22. With take-of faps set, Vx and Vy will be:

a. lower than that for clean confguration.b. higher than that for clean confguration.c. same as that for clean confguration.d. changed so that Vx increases and Vy decreases compared to clean confguration.23. The maximum rate of climb that can be maintained at the absolute ceiling is:a. 0 ft/minb. 125 ft/minc. 500 ft/mind. 100 ft/min24. A head wind will:a. increase the rate of climb.b. shorten the time of climb.c. increase the climb fight path angle.d. increase the angle of climb25. VX is:a. the speed for best rate of climb.b. the speed for best specifc range.c. the speed for best angle of fight path.d. the speed for best angle of climb.212Chapter 3General PrinciplesClimb and Descent26. The best rate of climb at a constant gross mass:a. decreases with increasing altitude since the thrust available decreases due to the lower air density.b. increases with increasing altitude since the drag decreases due to the lower air density.c. increases with increasing altitude due to the higher true airspeed.d. is independent of altitude.27. With a jet aeroplane, the maximum climb angle can be fown at approximately:a. 1.2 Vs.b. 1.1 Vs.c. The highest L/C ratio.d. The highest L/D ratio.28. During a climb with all engines operating, the altitude where the rate of climb reduces to 100 ft/min is called:

a. Thrust ceiling.b. Maximum transfer ceiling.c. Service ceiling.d. Absolute ceiling.29. With all other factors remaining constant, how does increasing altitude afect Vx and Vy as a TAS:

a. Vx will decrease and Vy will increase.b. Both will increase.c. Both will remain the same.d. Both will decrease.30. Any acceleration in climb, with a constant power seting:a. improves the climb gradient if the airspeed is below VX.b. improves the rate of climb if the airspeed is below VY.c. decreases rate of climb and increases angle of climb.d. decreases the rate of climb and the angle of climb.31. For an aircraft maintaining 100 kt true airspeed and a climb gradient of 3.3% with no wind, what would be the approximate rate of climb?a. 3.30 m/sb. 33.0 m/sc. 330 ft/mind. 3,300 ft/min32. During a climb to the cruising level, any headwind component:a. decreases the climb time.b. decreases the ground distance fown during that climb.c. increases the amount of fuel for the climb.d. increases the climb time.213Chapter 3General PrinciplesClimb and Descent33. The pilot of a single engine aircraft has established the climb performance. The carriage of an additional passenger will cause the climb performance to be:a. Degradedb. Improvedc. Unchangedd. Unchanged, if a short feld take-of is adopted.34. A headwind component increasing with altitude, as compared to zero wind condition: (assuming IAS is constant.a. improves angle and rate of climb.b. decreases angle and rate of climb.c. has no efect on rate of climb.d. does not have any efect on the angle of fight path during climb.35. Which of the following combinations adversely afects take-of and initial climb performance?a. High temperature and low relative humidity.b. Low temperature and low relative humidity.c. High temperature and high relative humidity.d. Low temperature and high relative humidity.36. A decrease in atmospheric pressure has, among other things, the following consequences on take-of performance:a. a reduced take-of distance and degraded initial climb performance.b. an increased take-of distance and degraded initial climb performance.c. a reduced take-of distance and improved initial climb performance.d. an increased take-of distance and improved initial climb performance.37. The angle of climb with faps extended, compared to that with faps retracted, will normally be:a. Increase at moderate fap seting, decrease at large fap seting.b. Smaller.c. Larger.d. Not change.38. What is the efect of tail wind on the time to climb to a given altitude?a. The time to climb increases.b. The time to climb decreases.c. The efect on time to climb will depend on the aeroplane type.d. The time to climb does not change.39. Changing the take-of fap seting from fap 15 to fap 5 will normally result in:a. a longer take-of distance and a beter climb.b. a shorter take-of distance and an equal climb.c. a beter climb and an equal take-of distance.d. a shorter take-of distance and a beter climb.214Chapter 3General PrinciplesClimb and Descent40. What is the infuence of the mass on maximum rate of climb (ROC. speed if all other parameters remain constant?a. The ROC is afected by the mass, but not the ROC speed.b. The ROC and the ROC speed are independent of the mass.c. The ROC speed increases with increasing mass.d. The ROC speed decreases with increasing mass.41. Following a take-of to the 50 ft (15 m. screen height, a light twin climbs on a gradient of 5%. It will clear a 160 m obstacle situated at 5,000 m from the 50 ft point with an obstacle clearance margin of:a. it will not clear the obstacle.b. 105 mc. 90 md. 75 m42. The climb gradient is defned as the ratio of:a. true airspeed to rate of climb.b. rate of climb to true airspeed.c. the increase of altitude to horizontal air distance expressed as a percentage.d. the horizontal air distance over the increase of altitude expressed as a percentage.43. When fying an aircraft at: i Vx without fap ii Vx with fap iii Vy without fap iv Vy with fapthe aircraft should be achieving:a. i The best rate of climb. ii The best rate of climb, but using- a slightly faster speed than in (i). iii The best angle of climb. iv The best angle of climb, but using a slightly faster speed than in (iii).b. i A good angle of climb. ii The best angle of climb. iii A good rate of climb. iv The best rate of climb.c. i The best angle of climb. ii A slightly reduced angle of climb compared to (i) if using a slightly reduced speed than in (i). iii The best rate of climb. iv A slightly reduced rate of climb compared to (iii) if using a slightly reduced speed than in (iii).d. i A good rate of climb.ii The best rate of climb.iii A good angle of climb.iv The best angle of climb.215Chapter 3General PrinciplesClimb and DescentDESCENT QUESTIONS1. Two identical aeroplanes at diferent masses are descending at idle thrust. Which of the following statements correctly describes their descent characteristics?a. At a given angle of atack, both the vertical and the forward speed are greater for the heavier aeroplane.b. There is no diference between the descent characteristics of the two aeroplanes.c. At a given angle of atack the heavier aeroplane will always glide further than the lighter aeroplane.d. At a given angle of atack the lighter aeroplane will always glide further than the heavier aeroplane.2. In a steady descending fight equilibrium of forces acting on the aeroplane is given by: (T = Thrust, D = Drag, W = Weight, descent angle = GAMMA)a. T + D = - W sin GAMMAb. T + W sin GAMMA = Dc. T - W sin GAMMA = Dd. T - D = W sin GAMMA3. Which of the following combinations has an efect on the angle of descent in a glide? (Ignore compressibility efects.) a. Confguration and mass.b. Confguration and angle of atack.c. Mass and altitude.d. Altitude and confguration.4. Which statement is correct for a descent without engine thrust at maximum lift to drag ratio speed?a. The mass of an aeroplane does not have any efect on the speed for descent.b. The higher the gross mass the greater is the speed for descent.c. The higher the gross mass the lower is the speed for descent.d. The higher the average temperature (OAT. the lower is the speed for descent.5. An aeroplane is in a power of glide at best gliding speed. If the pilot increases pitch atitude the glide distance:a. increases.b. remains the same.c. may increase or decrease depending on the aeroplane.d. decreases.6. Is there any diference between the vertical speed versus forward speed curves for two identical aeroplanes having diferent masses? (assume zero thrust and wind)a. Yes, the diference is that the lighter aeroplane will always glide a greater distance.b. Yes, the diference is that for a given angle of atack both the vertical and forward speeds of the heavier aeroplane will be larger.c. No diference.d. Yes, the diference is that the heavier aeroplane will always glide a greater distance.216Chapter 3General PrinciplesClimb and Descent7. Which statement is correct for a descent without engine thrust at maximum lift to drag ratio speed?a. A tailwind component increases fuel and time to descent.b. A tailwind component decreases the ground distance.c. A tailwind component increases the ground distance.d. A headwind component increases the ground distance.8. An aeroplane executes a steady glide at the speed for minimum glide angle. If the forward speed is kept constant at VMD, what is the efect of a lower mass on the Rate of descent / Glide angle / CL/CD ratio?a. decreases / constant / decreasesb. increases / increases / constantc. increases / constant / increasesd. decreases / constant/ constant9. Which of the following factors leads to the maximum fight time of a glide?a. Low mass.b. High mass.c. Headwind.d. Tailwind.10. A constant headwind:a. increases the descent distance over ground.b. increases the angle of the descent fight path.c. increases the angle of descent.d. increases the rate of descent.11. An aeroplane carries out a descent maintaining at a constant Mach number in the frst part of the descent and then at a constant indicated airspeed in the second part of the descent. How does the angle of descent change in the frst and in the second part of the descent? Assume idle thrust and clean confguration and ignore compressibility efects.a. Increases in the frst part; is constant in the second.b. Increases in the frst part; decreases in the second.c. Is constant in the frst part; decreases in the second.d. Decreases in the frst part; increases in the second.12. During a glide at constant Mach number, the pitch angle of the aeroplane will:a. decrease.b. increase.c. increase at frst and decrease later on.d. remain constant.13. Which of the following factors will lead to an increase of ground distance during a glide, while maintaining the appropriate minimum glide angle speed?a. Headwind.b. Tailwind.c. Increase of aircraft mass.d. Decrease of aircraft mass.217Chapter 3General PrinciplesClimb and Descent14. A twin jet aeroplane is in cruise, with one engine inoperative, and has to overfy a high terrain area. In order to allow the greatest height clearance , the appropriate airspeed must be the airspeed:a. giving the lowest Cd/Cl ratio.b. for long-range cruise.c. of greatest lift-to-drag ratio.d. giving the lowest Cl/Cd ratio.15. What is the efect of increased mass on the performance of a gliding aeroplane at VMD?a. The lift/drag ratio decreases.b. The speed for best angle of descent increases.c. There is no efect.d. The gliding angle decreases.16. A twin engined aeroplane in cruise fight with one engine inoperative has to fy over high ground. In order to maintain the highest possible altitude the pilot should choose:a. the speed corresponding to the minimum value of lift / drag ratio.b. the speed at the maximum lift.c. the speed corresponding to the maximum value of the lift / drag ratio.d. the long range speed.17. With all engines out, a pilot wants to fy for maximum time. Therefore he has to fy the speed corresponding to:a. the minimum power required.b. the critical Mach number.c. the minimum angle of descent.d. the maximum lift.18. Descending from cruising altitude to ground level at a constant IAS in a headwind, compared to still air conditions, will:a. Reduce the time to descend.b. Increase the time to descend.c. Reduce the ground distance taken.d. Reduce the fuel used in the descent.19. When descending at a constant Mach number:a. The angle of atack remains constant.b. The IAS decreases then increases.c. The pitch angle will increase.d. The pitch angle will decrease.218Chapter 3General PrinciplesClimb and DescentANSWERSCLIMB1 C 11 A 21 A 31 C 41 B2 B 12 C 22 A 32 B 42 C3 B 13 D 23 A 33 A 43 C4 C 14 A 24 C 34 C5 B 15 A 25 D 35 C6 C 16 C 26 A 36 B7 B 17 A 27 C 37 B8 A 18 A 28 C 38 D9 A 19 A 29 B 39 A10 C 20 D 30 D 40 CDESCENT1 A 11 A2 B 12 A3 B 13 B4 B 14 C5 D 15 B6 B 16 C7 C 17 A8 D 18 C9 A 19 D10 B219Chapter 4 General Principles - CruiseCHAPTER FOURGENERAL PRINCIPLES - CRUISEContentsBALANCE OF FORCES IN LEVEL FLIGHT . . . . . . . . . . . . . . . . . . . . . . 221MOVING THE CENTRE OF GRAVITY . . . . . . . . . . . . . . . . . . . . . . . . . 222AEROPLANE SPEEDS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223INDICATED AIRSPEED (IAS) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225CALIBRATED AIRSPEED (CAS) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226EQUIVALENT AIRSPEED (EAS) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226TRUE AIRSPEED (TAS) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226TRUE GOUNDSPEED (TGS) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227MACH NUMBER . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228ENDURANCE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230JET AEROPLANE ENDURANCE . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231PROPELLER AEROPLANE ENDURANCE. . . . . . . . . . . . . . . . . . . . . . . 231FACTORS AFFECTING ENDURANCE . . . . . . . . . . . . . . . . . . . . . . . . . 232RANGE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235FACTORS AFFECTING RANGE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238OPTIMUM ALTITUDE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245LONG RANGE CRUISE (LRC) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249QUESTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251ANSWERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258220Chapter 4 General Principles - Cruise221Chapter 4 General Principles - CruiseBALANCE OF FORCES IN LEVEL FLIGHTThis chapter will concentrate on the general performance principles of an aeroplane in the en-route phase of fight. Performance in the en route phase can be measured using the aeroplanes range and endurance parameters. These will be discussed in detail later on in the chapter. Let us frstly examine the forces acting on an aircraft in the cruise.These forces are split into couples and are shown in Figure 4.1. Figure 4.1 An illustration showing some of the forces in straight and level fightThe frst of these couples is produced by lift and weight. Weight acts through the centre of gravity of the aeroplane directly towards the centre of the earth. Lift balances weight and acts through the centre of pressure shown here. The efect of this couple on the aeroplane is to cause a nose down pitching moment. The lift weight couple is comparatively strong and thus the nose down pitching moment is large as shown by the large arrow pointing downwards to the right of the aeroplane. As an example of the size of these forces, in a 737-800 series, the maximum structural mass is 79,000 kg. At this mass the aeroplane weighs about 770,000 Newtons. Obviously this weight will be balanced in cruising fight by an equal and opposite lift force of 770,000 Newtons.The second couple acting upon an aeroplane in fight is that produced by the forces of thrust and drag. The efect of this couple is to cause a nose up pitching moment as shown by upward pointing red and green arrow to the right of the aeroplane. Notice though that this couple is far weaker than the lift weight couple. The maximum thrust produced by the engines of a 737-800 is only 214,000 Newtons. This means that the nose up pitching moment generated by the thrust drag couple does not balance the stronger nose down pitching moment of the lift weight couple. As a result there is still a nose down tendency as shown in Figure 4.1. In order to maintain level fight, we need somehow to generate an opposite moment which will balance the residual nose down pitching tendency. This is achieved by the tailplane, or horizontal stabilizer, on the aeroplanes tail assembly. The horizontal stabiliser, or tailplane must, be set at an angle which will cause a nose up pitching moment to balance the aeroplane, or more commonly expressed, to trim the aeroplane. Looking at Figure 4.2, notice that now, with the addition of the tailplane down force, the nose up and nose down pitching moments are in balance and level fight is possible. The down force generated at the tail is called tail plane down force or tail load. 222Chapter 4 General Principles - CruiseFigure 4.2 An illustration showing the complete balanceforces in straight and level fightAlthough such a force is necessary for level fight it does have two adverse efects on aircraft performance. Firstly, notice that the tail plane down force acts in the same direction as weight. It therefore increases the efective weight of the aircraft. The second adverse efect of the tailplane down force is its contribution to drag. The tail plane is an aerodynamic surface designed to produce lift. It will therefore produce induced drag as well as parasite drag. Therefore the greater the amount of balancing force produced by the tail plane the greater the aeroplane aerodynamic drag and efective weight. We will shortly understand that the two additional forces provided by the tail plane are detrimental to the aeroplanes en-route performance in terms of range and endurance.MOVING THE CENTRE OF GRAVITYTo some extent the amount of tail plane down force required for level fight can be manipulated by changing the centre of gravity. In fight this can be done in one of two ways. Firstly by selective fuel consumption, or secondly by fuel transfer. Consuming fuel in the tail frst, or transferring fuel out of the tail to other tanks, will move the centre of gravity position forewards. Conversely, using fuel in the centre or forward tanks frst, or transferring fuel out of these tanks will cause the centre of gravity to move aft. If the centre of gravity moves forwards, the magnitude of the lift/weight couple increases because the arm of the two forces is now longer. The greater strength from the lift weight couple increases the nose down pitching moment. This can be seen by comparing the length of the lift weight pitching down arrows from Figure 4.2 and Figure 4.3.223Chapter 4 General Principles - CruiseFigure 4.3 An illustration showing that with a more foreward C of G the lift weight couple increases and greater tail load is neededTo balance the greater nose down pitching moment and maintain level fight, more tail plane down force is needed. You may recall that this had the efect of increasing the weight and drag of the aeroplane which would have a detrimental efect on the aircrafts performance, reducing both its range and endurance. If the centre of gravity moves aft, the strength of the lift weight couple is decreased because the arm of the two forces is now shorter. This decreases the nose down pitching moment. To balance nose down pitching moment, less tail plane down force is needed to maintain level fight. Reducing tail plane down force reduces drag and efective weight which will increase both the aeroplanes range and endurance capability. Remember that, before you fy, when carrying out a weight and balance check, also ensure that the centre of gravity is still within the limits published in the aircraft manual. Be extra careful when handling data produced by countries whose units of measurement are diferent to those you are used to.AEROPLANE SPEEDSThis particular section of the chapter will deal with the aeroplanes speed, including; maximum speed, minimum speed, and the relationship between the various expressions of speed such as indicated airspeed, calibrated airspeed, true airspeed, true ground speed and mach number. Firstly let us examine what we mean by an aeroplanes maximum speed. You will have learnt from earlier chapters that an aeroplane will remain at a constant speed when the forward and rearward forces are balanced, as shown in Figure 4.4. 224Chapter 4 General Principles - CruiseFigure 4.4 In straight and level fight a constant speed is maintainedwhen the forward forces of thrust balances the rearward forces of dragIn this case, thrust is equal to drag. In order for the aeroplane to accelerate thrust must exceed drag. This can be achieved by the pilot opening the throtle further to increase power. With thrust greater than drag the aeroplane will accelerate. As the aeroplane accelerates, drag will increase. When drag reaches a value which is the same as the thrust, then acceleration will cease and the aeroplane will have achieved balanced fight once more, but now at a higher speed. Therefore the highest level fight speed that can be fown by the aeroplane will be at a speed where thrust is maximum and drag is maximum. Let us look at what we have just learnt but this time using the graphs. Shown in Figure 4.5 are the thrust and drag curves for a typical jet aeroplane. Figure 4.5 An illustration showing that that intersection of the thrust an drag curves represent the maximum and minimum straight and level fight speed225Chapter 4 General Principles - CruiseNotice the maximum speed is achieved once thrust and drag are equal. It is impossible in straight and level fight to accelerate any faster since the drag would exceed the thrust. This speed is the fastest speed the aeroplane can achieve in level fight. Look at the graph again and notice that there is another point on the curves where thrust and drag are equal. This point lies to the very left of the graph. The speed at this point represents the slowest speed that can be maintained. In level fight it would be impossible to fy any slower at this thrust seting since the rearward force of drag would exceed the force of thrust.

At high altitude the thrust produced by the engine decreases and therefore the green thrust line moves downwards.

Figure 4.6 An illustration showing that at high altitudes,the level fight speed range decreases.Looking at Figure 4.6 it is clear to see that at high altitude the maximum achievable level fight speed is slower and the minimum achievable level fight speed is faster than at lower altitudes, thus the range of speeds for the aeroplane is narrower.This next part of the chapter will focus on the explanation of the various expressions of aeroplane speed. INDICATED AIRSPEED (IAS)Diferent expressions of aeroplane speed are used for diferent purposes depending on whether we are concerned with aerodynamics, operations, navigation or even performance. In most cases aeroplane speed is measured using certain types of probes called the total pressure and static pressure probes. These probes help us to isolate dynamic pressure on which indicated airspeed is based. You can feel total pressure by simply puting your arm out of the window of a moving car. However, total pressure probes, sometimes called pitot probes and static probes, sufer from errors. Without any correction to the errors, the speed which is obtained from the probes when dynamic pressure is being sensed is called the Indicated airspeed, abbreviated to I A S. Indicated airspeed is the speed that is displayed on the airspeed indicator. 226Chapter 4 General Principles - CruiseCALIBRATED AIRSPEED (CAS)If the pressure probes are corrected for instrument and position errors the speed is called Calibrated Airspeed. Abbreviated to CAS. This speed is also known as rectifed airspeed, or RAS but this expression will not be used in this book. Calibrated airspeed is more accurate than indicated airspeed. In large modern commercial aeroplanes, the air data computer automatically corrects the pressure probe data for these errors and displays the Calibrated Airspeed in the airspeed indicator rather than the indicated airspeed. EQUIVALENT AIRSPEED (EAS)The fnal correction to make to the information received from the pressure probes is the correction to compensate for the efect of compressibility. Typically, at speeds beyond 220 knots, the air ahead of the aeroplane does not move out of the way. As a result, the air starts to build up and compress in front of the aeroplane. This build up of air has an efect called the compressibility efect. If the probes are corrected for the compressibility error as well as position and instrument errors, the speed obtained is called Equivalent airspeed, abbreviated to EAS. Equivalent airspeed is the most accurate of the speeds which are obtained from dynamic pressure. For the most part, this performance book will assume that indicated airspeed, calibrated airspeed and equivalent airspeed are the same. But, unless otherwise stated, assume any reference to aeroplane speed as being indicated airspeed.TRUE AIRSPEED (TAS)The next speed to consider is True Air Speed, abbreviated to TAS. True air speed is the Equivalent airspeed corrected for density error and as the name suggests is the true speed of the aircraft relative to the air through which the aeroplane is fying.Below is a very simplifed formula showing that true air speed is proportional to the equivalent airspeed and inversely proportional to the density. TAS is proportional to EAS DENSITYIf for a constant equivalent airspeed the density were to fall, the true air speed would increase. This means that with increasing altitude at a constant equivalent airspeed, true airspeed increases, as shown by Figure 4.7. Figure 4.7 An illustration showing the relationship between EAS and TAS with altitude227Chapter 4 General Principles - CruiseTrue airspeed can be calculated by using the tables in the aeroplane fight manual, using a fight navigation computer or even using a calibration scale on the airspeed indicator. True Airspeed is mainly used for navigation and fight planning purposes. TRUE GROUNDSPEED (TGS)The next speed to consider is the true ground speed, abbreviated to TGS or just GS. This represents the aeroplanes speed within a fxed ground reference system. Put more simply it is the aeroplanes velocity over the ground. The True Ground Speed is equal to the true airspeed plus or minus the wind component. If there is a tailwind then the aeroplanes speed over the ground will increase by a value equal to the speed of the tailwind. For example, if the true airspeed is 250 knots and the tailwind is 20 knots the true ground speed is 270 knots. Conversely, if there is a headwind, the aircrafts speed over the ground will be reduced by a value equal to the speed of the headwind.For example, if the true airspeed is 250 knots and the headwind is 30 knots the true ground speed is 220 knots. If the head wind or tailwind components are not already known, these can be worked out using a fight navigation computer or by using graphs or tables in the aeroplane fight manual. CAP 698, Section 4, Page 4, Figure 4.1 can also be used.At speeds higher than about 220 knots, some of the energy of the aeroplane goes into compressing the air ahead of the aeroplane and locally increasing the density of the air. Compressibility afects the amount of drag force on the aeroplane and the efect becomes more important as speed increases. As the aeroplane moves through the air it makes noise simply by disturbing the air. This noise emanates outwards in the form of pressure waves. These pressure waves stream out away from the aircraft at the speed of sound in all directions acting just like the ripples through water when a stone is dropped in the middle of a still pond.Figure 4.6 An illustration showing pressure waves emanating from atheoretically stationary aeroplaneHowever, as the aeroplane approaches the speed of sound it actually starts catching up with its own pressure waves in front if it as can be seen in Figure 4.7. These pressure waves turn into one big pressure shock wave which causes a loud bang, called a sonic boom. The shock wave generated actually bufets the aeroplane and decreases the lift force. 228Chapter 4 General Principles - CruiseFigure 4.7 An illustration showing pressure waves emanating froman aeroplane fying at the speed of soundObviously because of the increasing drag, decreasing lift and aeroplane bufet, pilots need to know when they are approaching the speed of sound. MACH NUMBERThe ratio of the speed of the aeroplane to the speed of sound in the air determines the magnitude and intensity of many of the efects of high speed fight. Because of the importance of this speed ratio, aerodynamicists have given it a special name called the Mach number in honour of Ernst Mach, a late 19th century physicist who studied gas dynamics. Mach number is not an actual speed as such; it is as we have already said the ratio of the true speed of the aeroplane to the local speed of sound. Mach number is best illustrated using an example. To calculate the Mach number, simply divide the true airspeed (TAS) by the local speed of sound (LSS). MACH NUMBER = TAS LSSAt sea level in ISA conditions, the local speed of sound is 661 knots. If the true airspeed of the aeroplane is 510 knots, then the Mach number is 0.77. In other words, the aeroplane is travelling at about three quarters of the speed of sound. If the aeroplane were to fy faster, the Mach number would increase. If the aeroplane accelerated to 661 knots, then its speed would be equal to the speed of sound and the aeroplane would be at Mach 1. Aeroplanes fying between 0.8 mach and 1.2 mach are said to be in transonic fight. However, the term super sonic is more commonly known. Super sonic fight describes fight speed above Mach 1.The speed of sound varies with temperature. As altitude increases, the reducing temperature causes the local speed of sound to fall. At 30,000ft the speed of sound is 590 knots. If the true airspeed of the aeroplane is kept constant at 510 knots, as a result of the falling local speed of sound with altitude, the Mach number will increase.As the aeroplanes speed approaches Mach 1, the compressibility and approaching shock wave can have very detrimental efects on the aeroplanes performance if the aeroplane is not designed to mitigate such efects. As a result, most commercial aeroplanes in service today have a limit on the maximum Mach number they are allowed to fy at. This maximum operating Mach number is called MMO. Having discussed all the relevant speeds of an aeroplane it is important to understand the relationship of these speeds with one another as altitude changes. This is best illustrated on a graph as shown in Figure 4.8.229Chapter 4 General Principles - CruiseFigure 4.8 An illustration showing the relationshipof the various speeds with altitude

If calibrated airspeed, or simply indicated airspeed is kept constant with altitude, then as density falls the true airspeed will increase, as shown in Figure 4.8. But if true airspeed increases with altitude, while the local speed of sound decreases, then the Mach number must increase. These lines representing these three speeds can be manipulated to help solve complex problems. Figure 4.9 An illustration showing the relationshipof the various speeds with altitudeIf the true airspeed were to be kept constant with altitude as shown in Figure 4.9, the TAS line would be drawn vertically and the CAS and Mach lines as shown. Now, calibrated airspeed decreases whilst Mach number increases.230Chapter 4 General Principles - CruiseIf Mach number were to be kept constant, the diagram could be drawn so that the mach line is straight up as illustrated in Figure 4.10. Notice that TAS and CAS decrease with increasing altitude. Figure 4.10 An illustration showing the relationshipof the various speeds with altitudeThese graphs can also be used to see the relationship between the speeds when descending, just ensure to follow the lines down not up. Therefore, looking at Figure 4.10, if the pilot was to descend at a constant Mach number, then the EAS and TAS will increase.To help you remember how to draw the lines, C, T and M always appear from left to right. You can use Britains favourite food Chicken Tikka Masala as an acronym for remembering C T and M and in the order to which they appear in the graph.ENDURANCEThe next section of this chapter analyses the two cruise performance parameters; range and endurance. When fying for range we are asking the question - how much fuel will the aeroplane use per unit distance? When fying for endurance we are asking the question - how much fuel does the aeroplane use per unit time? Let us frstly deal with the endurance of the aeroplane. The endurance of an aeroplane is the time it can remain airborne on a given quantity of fuel or, put another way, endurance can be expressed as fuel used over a given airborne time. The only time an aeroplane will be fown by the pilot for maximum endurance is when the aeroplane is in a holding patern over its destination. For instance, when there are long landing delays running out of fuel starts to become a problem.Endurance is defned to be the ratio of airborne time over fuel used for that time. ENDURANCE = TIME (hr) FUEL (kg)231Chapter 4 General Principles - CruiseHowever, to use this formula a small adjustment needs to be made as shown below. SPECIFIC ENDURANCE (hr/kg) = 1 FUEL FLOWThis now becomes the formula for specifc endurance. The units of Specifc Endurance are airborne hours per kilogram of fuel consumed. The endurance of an aeroplane is simply a function of its fuel fow or fuel consumption. An aeroplane which can minimise its fuel fow will achieve maximum endurance. Therefore, when analysing the maximum endurance capability of an aeroplane it is necessary to understand what controls the fuel fow. You should note that fuel fow calculations are diferent for a jet and propeller aeroplane. Let us examine the jet aeroplane frst. JET AEROPLANE ENDURANCEFUEL FLOW = FUEL FLOW PER UNIT THRUST X TOTAL THRUSTHere, fuel fow is a function of the fuel used per unit of thrust, multiplied by the total number of thrust units. Obviously if it were possible to reduce the fuel used per unit of thrust and the total number of thrust units required then the total fuel fow would be reduced. Fuel used per unit thrust is most commonly known as specifc fuel consumption which is abbreviated to SFC. The formulae for fuel fow should be as shown belowFUEL FLOW = SFC X TOTAL THRUSTThe specifc fuel consumption needs to be small, in other words, the aim is to reduce the amount of fuel used to produce each thrust unit. For a jet engine this occurs when ambient temperature is very low and engine rpm is very high. This can only occur at high altitude. So, fying at high altitude, minimises the fuel used per unit of thrust. Having minimised the fuel used to produce each unit of thrust, the aim is now to fy the aeroplane using minimum possible total thrust because each unit of thrust requires fuel to be consumed. This problem is comparatively simple to solve. In level fight the forward acting force of thrust is controlled and balanced by the rearward acting force of drag. If drag is small the aeroplane need fy with only a small amount of thrust. Looking at the formula below, thrust can be replaced by drag since the value of drag is equal and opposite to the value of thrust.FUEL FLOW = SFC X TOTAL DRAGTo minimise drag, the jet aeroplane simply fies at the velocity for minimum drag. Therefore VMD is the speed to fy for maximum endurance for a jet aeroplane In summary then, for a jet aeroplane to maximise its endurance by minimising its fuel fow, the pilot would fy the aeroplane at VMD and fy as high as possible.PROPELLER AEROPLANE ENDURANCEThe situation with endurance and fuel fow for a propeller aeroplane is very similar to that for the jet aeroplane. There is just one small diference in the formula we use. Turbo-prop and piston engines frst convert chemical energy in the fuel into power output on a shaft. The propeller then converts that power into thrust. Therefore, for propeller since the fuel is used to generate power and not thrust the formula for fuel fow is fuel used per unit of power multiplied by the total units of power. FUEL FLOW = FUEL FLOW PER UNIT POWER X TOTAL POWER232Chapter 4 General Principles - CruiseObviously to minimise the fuel fow, the fuel used per unit of power and the total number of power units must be kept to a minimum. Fuel used per unit power as you already know is called specifc fuel consumption. Therefore, similarly with a jet the formula for fuel fow for a propeller reads as below.FUEL FLOW = SFC X TOTAL POWERThe specifc fuel consumption value needs to be small but for the majority of propeller aeroplanes the value is more or less fxed. However, in very general terms it is safe to say that for piston engines specifc fuel consumption is a minimum at lower altitudes, whereas for turbo-propeller engines specifc fuel consumption is a minimum at middle to high altitudes. Since specifc fuel consumption is more or less fxed, the only other way to minimise fuel fow is to use the minimum amount of power. This can be achieved by fying at the speed for minimum power required, or VMP. Therefore note that, for a propeller aeroplane, it is VMP that is the speed for maximum endurance, whereas for a jet it is VMD.FACTORS AFFECTING ENDURANCEWEIGHTHaving studied how to achieve maximum endurance for both jet and propeller engines, let us now examine what factors actually afect endurance. The frst factor we must discuss is the efect of weight. You will recall that increasing the weight of the aeroplane increases induced drag and thus moves the total drag curve and power required curves up and right. Looking at Figure 4.11 you can see that for jet aeroplanes at higher weights, the aeroplane has more drag which will require more thrust and therefore require more fuel fow. In this situation the endurance will decrease. Figure 4.11 An illustration of the efect of weight on the drag and thespeed for maximum endurance for a jet aeroplane233Chapter 4 General Principles - CruiseWe must also note that the speed for maximum endurance, VMD is now higher. For a propeller aeroplane the situation is similar. For a propeller aeroplane at higher weights, more power is required, therefore fuel fow increases and endurance decreases, but also the speed for best endurance, VMP is higher. This can be seen in Figure 4.12.Figure 4.12 An illustration of the efect of weight on the power required and thespeed for maximum endurance for a propeller aeroplane

There is one other efect that is seldom mentioned. At higher weights, operating altitudes are lower, which for a jet aeroplane means that the specifc fuel consumption increases, because at lower altitudes, the jet engine is less efcient.CONFIGURATIONThe next factor which afects endurance is the aeroplanes confguration. While it seems an obvious point that the undercarriage and / or faps should not be deployed in the cruise, there will be occasions when a pilot will fnd himself stacked with other aeroplanes in a holding patern over the destination airfeld. As the aeroplane at the lowest level exits the hold to land, the other aeroplanes will have to descend to a lower hold and eventually prepare for the landing by deploying undercarriage and faps. You will recall that deploying the faps and undercarriage increases parasite drag and thus causes the total drag curve and power required curve to move up and left. Looking at Figure 4.13 which is for a jet aeroplane and uses the drag curve, you can see that with the gear and faps deployed, the aeroplane has more drag and therefore requires greater fuel fow, but notice that the speed for maximum endurance, VMD is now lower. 234Chapter 4 General Principles - CruiseFigure 4.13 An illustration of the efect of gear and faps on the drag and thespeed for maximum endurance for a jet aeroplaneUsing Figure 4.14 which is for a propeller aeroplane, it is much the same. With gear and faps deployed, more power is required, therefore fuel fow increases and endurance decreases, but also the speed for best endurance, VMP is lower. Figure 4.14 An illustration of the efect of gear and faps on the power required and thespeed for maximum endurance for a propeller aeroplaneFuel fow in the landing confguration can increase by 150% compared to a clean confguration, so it important not to deploy gear or faps too early and unnecessarily increase the fuel costs for the fight.235Chapter 4 General Principles - CruiseWIND AND ALTITUDEThe case of wind is quite simple. It has no efect on endurance. You will remember that maximum endurance is concerned with minimising fuel fow. It should be obvious that wind does not afect the fuel fow into the engine. Endurance is about time in the air, not distance covered. Whatever the efect of wind, an aeroplane will still remain airborne only for as long as it has useable fuel in its tanks. Altitude however, does afect endurance. Its efect though is a litle complicated and is very dependent on engine type. Generally jet aeroplanes become more efcient as altitude increases partly due to the decreasing ambient temperature, but also because of the increasing rpm required to maintain thrust. Therefore, theoretically, the maximum endurance of a jet aeroplane will be achieved when fying at or above the tropopause where the ambient air temperature is lowest. Turbo propeller aeroplanes function in a similar way to a jet since they are, in essence, jet engines with a propeller atached to a geared shaft. However, even though the turbo-propeller engine gains efciency with altitude, the power required increases due to the rising TAS ofseting the efciency gains. This means that for the majority of modern turbo-propeller aeroplanes, maximum endurance is achieved at around 10,000 ft or less.Piston engine aeroplanes are most efcient at sea level when the manifold pressure is high and rpm is low, provided that the mixture has been leaned correctly. In summary then, jet aeroplanes achieve maximum endurance at or above the tropopause; turbo propeller aeroplanes reach maximum endurance at about 10,000 ft and piston engine aeroplanes have their maximum endurance at sea level.RANGEYou have learnt that there are two performance parameters in the cruise; range and endurance. Lets us now consider range. Range is a more useful performance parameter than endurance, and one that aircraft designers continually try to improve. Whereas endurance is about airborne time, range is more concerned with distance covered and it is, therefore, sometimes referred to as fuel mileage. For range, not only is the concern to minimise the fuel fow, but more importantly to maximise the speed. This will allow the aeroplane to travel a greater distance.Maximum range can be defned as being the maximum distance an aeroplane can fy for a given fuel quantity consumed or to put it another way, the minimum fuel used by an aeroplane over a given distance. This later expression of range is more commonly used for commercial operations. As a formula, range is simply the distance in nautical miles divided by fuel quantity in kilograms. RANGE = DISTANCE (nm) FUEL (kg)However, in the same way as we did for the formula for endurance we must adjust our range formula needs in order for it to give us useful information. The range that an aeroplane can achieve is determined by the speed of the aeroplane and the fuel fow. The top line of the formula is nautical air miles per hour (TAS) and the botom line of the formula is kilograms of fuel per hour. Thus the formula now reads true air speed divided by fuel fow. SPECIFIC RANGE (SR) = TAS FUEL FLOW236Chapter 4 General Principles - CruiseThis is the formula for Specifc air range. The formula shows that specifc air range is defned as the ratio of true airspeed over the fuel fow. You may recall from the endurance section, that fuel fow is specifc fuel consumption multiplied by drag for a jet and specifc fuel consumption multiplied by power required for a propeller driven aeroplane. Looking at the formulae it is now obvious that in order to maximise the specifc range of the aeroplane, true air speed must be high, and the fuel fow must be low.JET SPECIFIC RANGE (SR) = TAS SFC X DRAGPROPELLER SPECIFIC RANGE (SR) = TAS SFC X POWER REQUIREDJET AEROPLANE RANGELet us now focus on the specifc air range of the jet aeroplane. We stated earlier that to maximise the range, the TAS must be high, and the specifc fuel consumption and the drag must be low. Shown in Figure 4.15 is a drag curve for a jet aeroplane. If the aeroplane were to fy at VMD then of course the drag force would be at its lowest. Referring to the formula for specifc range which shown in the graph it seems we have solved one of the points, namely how to make drag as low as possible. However, notice that because the drag curve is fairly fat at the botom, the speed may be increased signifcantly from VMD for only a small drag penalty.Figure 4.15 A graph showing the speed for maximum range for a jet aeroplaneWe can therefore see that whilst drag has increased a litle, which is bad for range, the air speed has increased signifcantly, which is good for range. Consequently the overall efect is that there is an increase in specifc range. 237Chapter 4 General Principles - CruiseThe speed at which the speed over drag ratio is maximised may be read from the graph at the tangent to the curve. You may recall that this speed was 1.32 VMD. Therefore it is 1.32 VMD that is the speed for maximum range for a jet aeroplane. There is now only one remaining item left in the formula which still needs to be resolved. In order to increase range even more, specifc fuel consumption must be decreased. You may recall that the only way to do this for a jet aeroplane is to operate at as high an altitude as possible. Operating as high as possible will give us a higher true airspeed for any given indicated airspeed which, again, will improve the specifc range.PROPELLER AEROPLANE RANGEHaving completed the analysis of range for a jet aeroplane, let us now examine specifc air range for a propeller aeroplane. We stated earlier that to maximise the specifc range, the TAS must be high, and the specifc fuel consumption and the power required must be low. Shown in Figure 4.16 is the power required curve for a propeller aeroplane. If the aeroplane were to fy at VMP then of course, its engine would be delivering minimum power required for level fight. It seems then, that we have solved one of the components, namely to make power required as small as possible.Figure 4.16 A graph showing the speed for maximum rangefor a propeller aeroplaneHowever, looking at the graph, you will notice, like for the jet aeroplane, that because the power required curve is fairly fat at the botom, the aeroplane speed may be increased signifcantly from VMP for only a small penalty increase in the power required. You can see that whilst power required has increased a litle, which is bad for range, the air speed has increased signifcantly, 238Chapter 4 General Principles - Cruisewhich is good for range. Consequently the overall efect is an increase in the range. The point at which the speed power ratio is at a maximum is the tangent point of the curve. You may recall that this speed is VMD. Therefore it is VMD that is the speed for maximum range for a propeller aeroplane. There is now only one remaining item left in the formula which still needs to be resolved. In order to maximise range even more, specifc fuel consumption must be decreased. However, you may remember that specifc fuel consumption for piston aeroplane is more or less best at low altitudes, where as for turbo-propeller aeroplanes which use jet engines, the specifc fuel consumption decreases with altitude up to a point about halfway up the troposphere.FACTORS AFFECTING RANGEWEIGHTHaving studied how to achieve maximum range for both jet and propeller engines, let us now examine what factors afect range. The frst of the factors to discuss is the efect of weight. You will recall that increasing the weight of the aeroplane increases induced drag and thus moves the total drag curve and power required curve up and right. Figure 4.17 A graph showing the efect of weight on drag and the speed formaximum range for a jet aeroplaneLooking at Figure 4.17 which is for a jet aeroplane, you can see for higher weights, the aeroplane is subject to a higher drag force and therefore requires a higher rate of fuel fow. This will decrease specifc range. However, notice that the speed for maximum range, 1.32 VMD is now higher. 239Chapter 4 General Principles - CruiseFigure 4.18 A graph showing the efect of weight on power required and thespeed for maximum range for a propeller aeroplane.Using Figure 4.18 which is for a propeller driven aeroplane, it is much the same as the jet. At higher weights, more power is required, therefore fuel fow increases and range decreases. However, notice too, that the speed for best range, VMD, is higher.Remember also that at higher weights, the operating altitudes are reduced, which for a jet aeroplane means that the specifc fuel consumption increases, because at lower altitudes, the jet engine is less efcient. 240Chapter 4 General Principles - CruisePAYLOAD Vs RANGEOne of the most important issues that airline operators need to consider is the choice of the aeroplane they operate. This choice is mainly based upon the required aeroplanes payload and range. These requirements can be best described by going through a typical example. Shown in Figure 4.19 is the payload range graph for a Boeing triple seven. On the vertical axis is the payload in thousands of kilograms and the horizontal axis shows the aeroplane range in thousands of nautical miles.Figure 4.19 A graph showing the payload verses therange of a typical Boeing 777Using Figure 4.19 as the payload is initially added to the aeroplane the marker moves from point A to point B. However, payload will reach a maximum when either there is no more space on the aeroplane or the aeroplane has reached its zero fuel mass (ZFM). Notice that the range at point B is zero because no fuel has been added yet. Fuel is now added to the aeroplane and the blue marker line moves to the right showing an increase in range. Adding fuel can continue until the maximum structural take of mass is achieved. This is shown by point C on the graph. Although maximum mass has been reached, it is unlikely that the tanks are full at this stage. To increase range further, more fuel must be added. But since the maximum mass has been achieved, the only way to add more fuel is if some of the payload is exchanged for fuel. Now the marker will start to move down and right. This shows that range is increasing but the payload is decreasing. Swapping payload for fuel in this way can continue only until the tanks are full, as shown by point D. From point C to point D the total mass of the aeroplane has remained constant since we are simply swapping payload for fuel.The only way to increase the range beyond point D, even though the tanks are full, is to remove the rest of the payload. You will recall that reducing weight increases range. Reducing the payload completely moves the marker line from point D to point E. At point E, the aeroplane has full tanks, maximum range but no payload. In the majority of airlines the trade of between range and payload is carried out on initial aeroplane purchase and thereafter, during in fight planning. As a pilot it is unlikely you will be required to work through these graphs other than to check and confrm the data that has been prepared in advanced for you.241Chapter 4 General Principles - CruiseCONFIGURATIONAnother factor afecting the range is the aeroplanes confguration. You will recall that deploying the faps and undercarriage increases parasite drag and thus moves the total drag curve and power required curve up and left. Figure 4.20 A graph showing the efect of confguration on drag and the speed formaximum range for a jet aeroplaneShown in Figure 4.20 is the graph for the jet aeroplane which is up and to the left using the drag curve. You can see that with the gear and faps deployed, the aeroplane has more drag and therefore needs more thrust and which, in turn, requires greater fuel fow. This will decrease the range. However, notice that the speed for maximum range, 1.32 VMD is now lower.242Chapter 4 General Principles - CruiseFigure 4.21 A graph showing the efect of confguration on power required and thespeed for maximum range for a propeller aeroplaneLooking at Figure 4.21 which is the range graph for a propeller aeroplane you can see that maters are much the same. With gear and faps deployed, more power is required. Therefore fuel fow increases and range decreases. But also notice that the speed for best range, VMD is lower.A point always to bear in mind in the cruise is that any increase in parasite drag will be detrimental to range and endurance. Increases in parasite drag can be caused by any number of things such as damaged and misaligned surfaces. The extra drag created by misaligned or misrigged airframe surfaces creates a type of drag called excrescence drag. This can be more than 4% of the aeroplanes total drag. Careful pre-fight inspection should reveal misaligned or misrigged surfaces.An important point to consider over which the pilot has direct control is aeroplane trim. A pilot must periodically check the aileron and rudder trim, the spoiler misfair and the trailing edges to ensure that the aeroplane is in trim and is in balanced fight. Monitoring the aeroplanes control surfaces will help to reduce the extra drag and extra fuel consumption that out of trim and unbalanced fight can cause.Lastly, contamination on the airframe from airframe icing can afect fuel fow. This icing will not only change the shape of the wing, making it less efcient at producing lift, but it will also increase weight and drag. All this is detrimental to aircraft performance and will reduce the aeroplanes range.243Chapter 4 General Principles - CruiseWINDWind is the next of the major factors afecting the range of an aeroplane. Headwinds will cause the aircraft to travel slower over the ground and therefore cover less distance for a given level of fuel consumption. Thus in headwinds range is reduced. In order to minimise this efect the speed of the aeroplane is increased by a margin slightly less than the amount of the headwind. The increase in speed will increase the thrust and the power required and therefore the fuel consumption, but on a positive side the aeroplane will be exposed to the headwind for a shorter time period if it is fying faster. This higher speed then recovers some of the range loss caused by the headwind. On the contrary, a tailwind will increase the ground speed and increase the distance covered for a given level of fuel consumption, thereby increasing range. For maximum range with a tailwind, the speed for the best range should be decreased slightly so as to reduce the thrust and power required. This will therefore reduce the fuel fow and increase the range a litle more. The reduction in speed is slightly less than the speed of the tailwind component being experienced.Figure 4.22 A graph showing the efect of a headwind on the speed formaximum range for a jet aeroplane.The headwind and tailwind speed changes can be seen on the drag and powers curves for the jet and propeller aeroplane. However, to simplify things, well just concentrate on the drag curve for the jet aeroplane.Looking at Figure 4.22, with a 20 knot headwind, the origin of the tangent line moves 20 knots to the right. Notice that the tangent meets the curve at a point corresponding to a higher speed, thus confrming that in a headwind the speed for best range is higher. The opposite is the case in a tailwind scenario. With a tailwind of 20 knots, the origin of the tangent line moves 20 knots left. The tangent meets the curve at a point corresponding to a lower speed. This confrms that in a tailwind the speed for best range is lower than in conditions of zero wind.244Chapter 4 General Principles - CruiseALTITUDE (JET AEROPLANES)The efect of altitude on the range of an aeroplane is important, especially for a jet aeroplane. Below is the formula for the specifc air range for a jet aeroplane. Let us examine how the variables change with increasing altitude. SPECIFIC RANGE (SR) = TAS / SFC X DRAGAs aeroplane operating altitude increases, the colder air and requirement for increasing rpm cause the specifc fuel consumption to decrease which will help to increase the specifc air range. However there are two other variables left to consider, namely the true airspeed and the drag. Lets deal with true air speed frst. If you think back to the part of the lesson where we were analysing the efects of altitude on the various speeds, you will recall that if the aeroplane operates at higher and higher altitudes at the constant indicated or calibrated airspeed of 1.32 VMD, the true airspeed increases. The efect of increasing altitude and therefore increasing true airspeed acts together with the reducing specifc fuel consumption to help increase the specifc range. Therefore, specifc range increases with altitude. However, the last element of the formula to consider is drag. You will recall that with altitude, as the true airspeed increases and the local speed of sound decreases, the Mach number increases. This means that the aeroplane is approaching the speed of sound and approaching its maximum operating mach number, MMO. The problem with this is that beyond a certain mach number, the compressibility factor and approaching shockwave will cause drag to increase. This will be detrimental to the specifc range as you can see on the formula. However, it is a litle more complicated. As altitude increases, if the Mach number is allowed to get too high, the penalty due to drag will start to outweigh the benefts of increasing TAS and reducing specifc fuel consumption. It is at this point that the specifc air range will start to reduce. Using the left hand blue line in the graph in Figure 4.23 you can see that initially specifc range increases with altitude but then above a certain altitude the specifc range decreases.Figure 4.23 A graph showing the efect of altitude on specifc range for ajet aeroplane at high and low weights245Chapter 4 General Principles - CruiseOPTIMUM ALTITUDEUsing the left hand blue line, in Figure 4.23, you will notice that there is an altitude at which the specifc range is greatest; in our example this is just below 33,000 ft. This altitude is called the Optimum Altitude. It is defned as being the pressure altitude which provides the greatest specifc range or fuel mileage at a given weight and speed. Flying higher or lower than the optimum altitude will decrease the range of the aeroplane.It is important to understand that the optimum altitude is not fxed. You will recall that as the weight decreases through fuel burn, the drag curve moves down and left. Therefore, the best range speed, 1.32 VMD, falls and the total drag decreases. Therefore, with decreasing weight the aeroplane needs to slow down to maintain the best range speed. As it does so, the Mach number will also decrease meaning that the aeroplane is not limited by the high Mach number and corresponding high drag. This fact allows the aeroplane to climb a litle. As the aeroplane climbs the Mach number will increase again to its previous limiting value and drag will increase back to its previous value. But more importantly the higher altitude has decreased the specifc fuel consumption. Therefore, the specifc air range increases during this litle climb. This means that over time, as the weight decreases with fuel burn, the optimum altitude increases. You can see this in Figure 4.23 by comparing the specifc range line for high and low weight. Notice too that the as the optimum increases the specifc range increases. Ploting the change of the optimum altitude over time can be seen in Figure 4.24.Figure 4.24 A graph showing the optimum altitude increasing with height as thefight progresses for a typical jet aeroplaneIn order for the aeroplane to maximise the specifc range the aeroplane must stay with the optimum altitude as the optimum altitude slowly increases, in other words the aeroplane must climb along the green line shown in Figure 4.23. Climbing in this way is sometimes called a cruise climb. But carrying out a cruise climb is not always possible, since air trafc control and airspace congestion may predetermine fight cruising levels. If this is the case, in order to stay close to the optimum altitude, step climbs may be performed and are shown by the dashed yellow line in Figure 4.24. 246Chapter 4 General Principles - CruiseSTEP CLIMBSStep climbs essentially mean that the aeroplane climbs to about 2000 ft above the optimum altitude and levels of. As fuel is used and weight falls, the optimum altitude will increase to a point where it is again 2000 ft above the aeroplanes current level but it can take up to 3 hours for it to do so. At its current level the aeroplane can then climb 4000 ft and level of so that it will once again be 2000ft above the optimum altitude. But, if the last step climb is within 200 nm of the top of descent, then the fuel saving is negated and the aeroplane should remain level.The step climb process can be repeated throughout the cruise and it helps to explain why the cruise altitudes at the end of the fight are higher than at the start. Carrying out step climbs in this way, rather than always staying with the optimum altitude will increase fuel consumption by about 1% and therefore decrease the maximum range by 1%. This may not sound like much, but over a year a typical 747 would have used an extra 34,000 tonnes of fuel. If an aeroplane did not even step climb and simply remained at a constant altitude during the cruise, then the aeroplane would increase its fuel consumption by 10% compared to fying constantly at the optimum altitude. This demonstrates just how important altitude and speed control is in the cruise for a typical commercial fight. ALTITUDE (PROPELLER AEROPLANES)Having discussed the efect of altitude on the jet aeroplane, let us now consider the efect of altitude on the propeller aeroplane. In general though, we may say that most turbo propeller aeroplanes operate signifcantly lower than their jet counterparts. Turbo propeller aeroplanes seldom operate above 30,000 feet, and therefore never really sufer from the efects of geting close to the speed of sound. Turbo-propellers are based on the same engine design as a pure jet, therefore, the efect of altitude on the turbo propeller is very similar to the jet aeroplane. As altitude increases the increasing TAS and slightly decreasing specifc fuel consumption help to improve the specifc range. However, this beneft is ofset a litle by the increasing power required at higher altitude. So whilst specifc range does improve with altitude, above 10,000ft it only improves by a small amount. The choice of altitude may depend more on the wind considerations and the time and fuel considerations involved in climbing to the selected altitude.The other type of propeller aeroplane is that driven by a piston engine. You will recall that the piston engine aeroplane has a more or less fxed specifc fuel consumption even though specifc fuel consumption is lowest at high manifold pressures, low rpm and with the mixture correctly set. Therefore the only remaining variables in the specifc air range formula for the piston engine aeroplane are the true airspeed and the power required. 247Chapter 4 General Principles - CruiseFigure 4.25 A graph showing the specifc range with altitude for atypical piston propeller driven aeroplaneAs the aeroplane operates at higher and higher altitudes the power required to maintain the range speed will increase. This of course is detrimental to range. However, as altitude increases, true airspeed increases for any given indicated airspeed and this is good for range. This fact slightly more than ofsets the increase of power required and therefore the specifc range slowly increases with altitude. However, the aeroplane will reach an altitude where the throtle needs to be fully advanced to maintain the selected speed. This altitude is called the full throtle height and it is shown in Figure 4.25. Beyond this altitude the selected power and selected airspeed cannot be maintained and the aeroplane will slow down. Very soon after this altitude, the true airspeed will also start to fall despite the decreasing density. This fact, combined with the constantly increasing amount of power required means that the specifc range will decrease. As a result, maximum specifc range will be atained just after full throtle height.WIND ALTITUDE TRADE-OFFThe efect of headwinds and tailwinds on the range of the aeroplane can play a signifcant role in the choice of cruising altitude. For example, if there is a considerable headwind at the selected cruising altitude, this will be detrimental to the range. In this case, it may be benefcial to operate at a diferent altitude where the winds might be more favourable. In large commercial operations most of these considerations are dealt with prior to the fight by the fight planning personnel. However, in smaller operations, and if conditions change in fight, a pilot may have to carry out a wind altitude trade of calculation. Information enabling the pilot to do this is usually given in the aeroplane fight manual, an example of which is shown in Figure 4.26. 248Chapter 4 General Principles - CruiseFigure 4.26 An illustration of a typical wind altitude trade of graphfor an Airbus aeroplane249Chapter 4 General Principles - CruiseLONG RANGE CRUISE (LRC)Figure 4.27 shows the relationship between the speed of the aeroplane and its range. The top of the blue curve represents the point of maximum range and the speed to which this is found. For a jet aeroplane you will recall this speed is 1.32 VMD. In commercial operations this speed is more commonly referred to as the Maximum Range Cruise or MRC. Figure 4.27 A graph showing the relationship of aeroplane speed and rangeHowever, maximum range cruise speed is seldom fown. Usually, a higher speed is used. Looking at the top of the graph, you will notice the line is fairly fat. This means that a signifcant speed increase can be achieved with only a small compromise in range. This higher speed is called the Long Range Cruise (LRC). The Long Range Cruise speed is about 4% higher than the Maximum Range Cruise speed, and as such the specifc range reduces by about 1%. The reason for using this higher speed is simply that there are costs other than fuel which need to be considered in commercial operations. A more detailed explanation of the relationships of these costs and how they afect the operations will be dealt with later on.250Chapter 4 General Principles - Cruise251Chapter 4 General Principles - CruiseQUESTIONS1. On a reciprocating engined aeroplane, with increasing altitude at constant gross mass, constant angle of atack and confguration, the power required:a. remains unchanged but the TAS increases.b. increases and the TAS increases by the same percentage.c. increases but TAS remains constant.d. decreases slightly because of the lower air density.2. Moving the centre of gravity from the forward to the aft limit: (gross mass, altitude and airspeed remain unchanged)a. increases the power required.b. afects neither drag nor power required.c. increases the induced drag.d. decreases the induced drag and reduces the power required.3. For jet-engined aeroplanes operating below the optimum altitude, what is the efect of increased altitude on specifc range?a. It does not change.b. Increases only if there is no wind.c. Increases.d. Decreases.4. If the thrust available exceeds the thrust required for level fight:a. the aeroplane accelerates if the altitude is maintained.b. the aeroplane descends if the airspeed is maintained.c. the aeroplane decelerates if it is in the region of reversed command.d. the aeroplane decelerates if the altitude is maintained.5. Given a jet aircraft. Which order of speeds is correct?a. Vs, Maximum range speed, Vx,b. Maximum endurance speed, Maximum range speed, VX.c. Vs, Vx, Maximum range speed.d. Maximum endurance speed, Long range speed, Maximum range speed.6. The pilot of a light twin engine aircraft has calculated a 4,000 m service ceiling with a take-of mass of 3,250 kg, based on the general forecast conditions for the fight. If the take-of mass is 3,000 kg, the service ceiling will be:a. less than 4,000 m.b. unchanged, equal to 4,000 m.c. only a new performance analysis will determine if the service ceiling is higher or lower than 4,000m.d. higher than 4,000 m.252Chapter 4Circling Approach 7. Consider the graphic representation of the power required for a jet aeroplane versus true air speed (TAS). When drawing the tangent out of the origin, the point of contact determines the speed of:

a. critical angle of atack.b. maximum endurance.c. minimum power.d. maximum specifc range.8. In the drag versus speed curve for a jet aeroplane, the speed for maximum range corresponds with:a. the point of contact of the tangent from the origin to the drag curve.b. the point of intersection of the parasite drag curve and the induced drag curve.c. the point of contact of the tangent from the origin to the parasite drag curve.d. the point of contact of the tangent from the origin to the induced drag curve.9. The speed VS is defned as the:a. speed for best specifc range.b. stalling speed or minimum steady fight speed at which the aeroplane is controllable.c. safety speed for take-of in case of a contaminated runway.d. design stress speed.10. What is the efect of a head wind component, compared to still air, on the maximum range speed (IAS) and the speed for maximum climb angle respectively?a. Maximum range speed decreases and maximum climb angle speed decreases.b. Maximum range speed increases and maximum climb angle speed increases.c. Maximum range speed increases and maximum climb angle speed stays constant.d. Maximum range speed decreases and maximum climb angle speed increases.11. A jet aeroplane is fying at the long range cruise speed at the optimum altitude. How does the specifc range / fuel fow change over a given time period?a. Decrease / decrease.b. Increase / decrease.c. Increase / increase.d. Decrease / increase.12. The maximum indicated air speed of a piston engined aeroplane, in level fight, is reached:a. at the service ceiling.b. at the practical ceiling.c. at the lowest possible altitude.d. at the optimum cruise altitude.13. The optimum cruise altitude increases:a. if the aeroplane mass is decreased.b. if the temperature (OAT) is increased.c. if the tailwind component is decreased.d. if the aeroplane mass is increased.253Chapter 4 General Principles - Cruise14. What afect has a tailwind on the maximum endurance speed?a. No afect.b. Tailwind only afects holding speed.c. The IAS will be increased.d. The IAS will be decreased.15. Which of the equations below defnes specifc air range (SR)?a. SR = Groundspeed/Total Fuel Flowb. SR = True Airspeed/Total Fuel Flowc. SR = Indicated Airspeed/Total Fuel Flowd. SR = Mach Number/Total Fuel Flow16. Which of the following statements, with regard to the optimum altitude (best fuel mileage), is correct?a. An aeroplane usually fies above the optimum cruise altitude, as this provides the largest specifc range.b. An aeroplane sometimes fies above the optimum cruise altitude, because ATC normally does not allow to fy continuously at the optimum cruise altitude.c. An aeroplane always fies below the optimum cruise altitude, as otherwise Mach bufet can occur.d. An aeroplane always fies on the optimum cruise altitude, because this is most atractive from an economy point of view.17. The optimum altitude:a. is the altitude up to which cabin pressure of 8 000 ft can be maintained.b. increases as mass decreases and is the altitude at which the specifc range reaches its maximum.c. decreases as mass decreases.d. is the altitude at which the specifc range reaches its minimum.18. A lower airspeed at constant mass and altitude requires:a. less thrust and a lower coefcient of lift.b. more thrust and a lower coefcient of lift.c. more thrust and a lower coefcient of drag.d. a higher coefcient of lift.19. The point at which a tangent out of the origin touches the power required curve:a. is the point where Drag coefcient is a minimum.b. is the point where the Lift to Drag ratio is a minimum.c. is the maximum drag speed.d. is the point where the Lift to Drag ratio is a maximum.20. The long range cruise speed (LRC) is in relation to the speed for maximum range cruise (MRC)?a. Lower.b. Depending on the OAT and net mass.c. Depending on density altitude and mass.d. Higher254Chapter 4Circling Approach 21. The maximum horizontal speed occurs when:a. The thrust is equal to minimum drag.b. The thrust does not increase further with increasing speed.c. The maximum thrust is equal to the total drag.d. The thrust is equal to the maximum drag.22. Under which condition should you fy considerably lower (4,000 ft or more) than the optimum altitude?a. If at the lower altitude either more headwind or less tailwind can be expected.b. If at the lower altitude either considerably less headwind or considerably more tailwind can be expected.c. If the maximum altitude is below the optimum altitude.d. If the temperature is lower at the low altitude (high altitude inversion. .23. The optimum cruise altitude is:a. the pressure altitude up to which a cabin altitude of 8,000 ft can be maintained.b. the pressure altitude at which the TAS for high speed bufet is a maximum.c. the pressure altitude at which the best specifc range can be achieved.d. the pressure altitude at which the fuel fow is a maximum.24. Maximum endurance for a piston engined aeroplane is achieved at:a. The speed that approximately corresponds to the maximum rate of climb speed.b. The speed for maximum lift coefcient.c. The speed for minimum drag.d. The speed that corresponds to the speed for maximum climb angle.25. On a long distance fight the gross mass decreases continuously as a consequence of the fuel consumption. The result is:a. The speed must be increased to compensate the lower mass.b. The specifc range increases and the optimum altitude decreases.c. The specifc range decreases and the optimum altitude increases.d. The specifc range and the optimum altitude increases.26. A jet aeroplane is climbing at constant Mach number below the tropopause. Which of the following statements is correct?a. IAS decreases and TAS decreases.b. IAS increases and TAS increases.c. IAS decreases and TAS increases.d. IAS increases and TAS decreases.27. Why are step climbs used on long distance fights?a. Step climbs do not have any special purpose for jet aeroplanes; they are used for piston engine aeroplanes only.b. ATC do not permit cruise climbs.c. To fy as close as possible to the optimum altitude as aeroplane mass reduces.d. Step climbs are only justifed if at the higher altitude less headwind or more tailwind can be expected.255Chapter 4 General Principles - Cruise28. Which of the following sequences of speed for a jet aeroplane is correct? (from low to high speeds. a. Maximum endurance speed, maximum range speed, maximum angle of climb speed.b. Maximum endurance speed, long range speed, maximum range speed.c. Vs, maximum angle climb speed, maximum range speed.d. Vs, maximum range speed, maximum angle climb speed.29. The pilot of a jet aeroplane wants to use a minimum amount of fuel between two airfelds. Which fight procedure should the pilot fy?a. Maximum endurance.b. Holding.c. Long range.d. Maximum range.30. Long range cruise is selected as:

a. the higher speed to achieve 99% of maximum specifc range in zero wind.b. the speed for best economy (ECON)c. the climbing cruise with one or two engines inoperative.d. specifc range with tailwind.31. The optimum long-range cruise altitude for a turbo-jet aeroplane:a. is only dependent on the outside air temperature.b. increases when the aeroplane mass decreases. c. is always equal to the powerplant ceiling.d. is independent of the aeroplane mass.32. Maximum endurance:a. is the same as maximum specifc range with wind correction.b. can be fown in a steady climb only.c. can be reached with the best rate of climb speed in level fight.d. is achieved in unaccelerated level fight with minimum fuel consumption.33. For a piston engined aeroplane, the speed for maximum range is:a. that which gives the maximum lift to drag ratio.b. that which gives the minimum value of power.c. that which gives the maximum value of lift.d. 1.4 times the stall speed in clean confguration.34. The speed for maximum endurance:a. is always higher than the speed for maximum specifc range.b. is always lower than the speed for maximum specifc range.c. is the lower speed to achieve 99% of maximum specifc range.d. can either be higher or lower than the speed for maximum specifc range.256Chapter 4Circling Approach 35. The intersections of the thrust available and the drag curve are the operating points of the aeroplane:a. in un-accelerated climb.b. in un-accelerated level fight.c. in descent with constant IAS.d. in accelerated level fight.36. For a jet transport aeroplane, which of the following is the reason for the use of maximum range speed?a. Minimises specifc fuel consumption.b. Minimises fuel fow.c. Longest fight duration.d. Minimises drag.37. The centre of gravity moving near to, but still within, the aft limit;a. increases the stalling speed.b. improves the longitudinal stability.c. decreases the maximum range.d. improves the maximum range.38. A jet aeroplane is performing a maximum range fight. The speed corresponds to:a. the minimum drag.b. the minimum required power.c. the point of contact of the tangent from the origin to the power required (Pr. versus TAS curve.d. the point of contact of the tangent from the origin to the Drag versus TAS curve.39. During a cruise fight of a jet aeroplane at a constant fight level and at the maximum range speed, the IAS / the drag will:a. increase / increase.b. decrease / increase.c. decrease / decrease.d. increase / decrease.40. Which of the following is a reason to operate an aeroplane at long range speed?a. The aircraft can be operated close to the bufet onset speed.b. In order to prevent loss of speed stability and tuck-under.c. It ofers greatly reduced time costs than with maximum range speed.d. In order to achieve speed stability.41. Long range cruise is a fight procedure which gives:a. an IAS which is 1% higher than the IAS for maximum specifc range.b. a specifc range which is 99% of maximum specifc range and a lower cruise speed.c. a specifc range which is about 99% of maximum specifc range and higher cruise speed.d. a 1% higher TAS for maximum specifc range.257Chapter 4 General Principles - Cruise42. The lowest point of the drag or thrust required curve of a jet aeroplane, is the point for:a. minimum drag and maximum endurance.b. maximum specifc range and minimum power.c. minimum power.d. minimum specifc range.43. If other factors are unchanged, the fuel mileage or range (nautical miles per kg. is:a. independent from the centre of gravity position.b. lower with an aft centre of gravity position.c. higher with a forward centre of gravity position.d. lower with a forward centre of gravity position.44. To achieve the maximum range over ground with headwind the airspeed should be:a. lower compared to the speed for maximum range cruise with no wind.b. reduced to the gust penetration speed.c. higher compared to the speed for maximum range cruise with no wind.d. equal to the speed for maximum range cruise with no wind.45. When utilising the step climb technique, one should wait for the weight reduction, from fuel burn, to result in:a. The aerodynamic ceiling to increase by approximately 2,000ft above the present altitude, whereby one would climb approximately 4,000ft higher.b. The optimum altitude to increase by approximately 2,000ft above the present altitude, whereby one would climb approximately 4,000ft higher.c. The manoeuvre ceiling to increase by approximately 2,000ft above the present altitude, whereby one would climb approximately 4,000ft higher.d. The en-route ceiling to increase by approximately 2,000ft above the present altitude, whereby one would climb approximately 4,000ft higher.46. Which of the following statements is true regarding the performance of an aeroplane in level fight?a. The maximum level fight speed will be obtained when the power required equals the maximum power available from the engine.b. The minimum level fight speed will be obtained when the power required equals the maximum power available from the engine.c. The maximum level fight speed will be obtained when the power required equals the minimum power available from the engine.d. The maximum level fight speed will be obtained when the power required equals the power available from the engine.258Chapter 4Circling Approach ANSWERS1 B 21 C 41 C2 D 22 B 42 A3 C 23 C 43 D4 A 24 D 44 C5 C 25 D 45 B6 D 26 A 46 A7 B 27 C8 A 28 C9 B 29 D10 C 30 A11 B 31 B12 C 32 D13 A 33 A14 A 34 B15 B 35 B16 B 36 B17 B 37 D18 D 38 D19 D 39 C20 D 40 C259Chapter 5 General Principles - LandingCHAPTER FIVEGENERAL PRINCIPLES - LANDINGContentsLANDING DISTANCE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261LANDING DISTANCE AVAILABLE (LDA). . . . . . . . . . . . . . . . . . . . . . . . 262LIFT AND WEIGHT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262REVERSE THRUST . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263DRAG . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264LANDING DISTANCE FORMULA . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266EFFECT OF VARIABLE FACTORS ON LANDING DISTANCE . . . . . . . . . . . . 267HYDROPLANING. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271LANDING TECHNIQUE ON SLIPPERY RUNWAYS . . . . . . . . . . . . . . . . . . 272MICROBUSTS AND WINDSHEAR . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273QUESTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275ANSWERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278260Chapter 5 General Principles - Landing261Chapter 5 General Principles - LandingLANDING DISTANCEThe landing stage of fight is defned as being that stage of fight commencing from 50ft above the landing threshold and terminating when the aeroplane comes to a complete stop as shown by Figure 5.1. The 50 ft point is sometimes referred to as the landing screen height. The landing screen height is fxed at 50 ft for all classes of aeroplane unlike the take-of screen height which is 35 ft for class A aeroplanes and 50 ft for class B aeroplanes.From the approach down to the landing screen height the aeroplane must have atained the landing reference speed, known as VREF. VREF for Class A aeroplanes must be no less than the greater of 1.23 times the stall reference speed in the landing confguration (1.23 VSRO) and the velocity of minimum control in the landing confguration (VMCL). V REF for all other classes of aeroplane must be no less than 1.3 times the stall speed (1.3 VSO) in the landing confguration. VREF is a very important speed to atain since the landing distances in the aeroplane fight manual are based on aeroplanes fying at VREF. Therefore if a landing aeroplane is not at VREF the landing distance given by the manual will not be achieved by the pilot. A landing carried out at a speed other than VREF could seriously jeopardize the safety of the landing. Figure 5.1 Landing DistanceFigure 5.1 Landing DistanceThe landing can be divided into two parts. We call these the airborne section and the ground run or landing roll The frst part, the airborne section starts from the landing screen height of 50 feet and ends when the aeroplanes main wheels touch the landing surface. The airborne section is usually given as being about 1000ft in length. Within the airborne section certain critical actions take place. On descending through the screen height the thrust is reduced to zero and the aeroplane pitch atitude is increased slightly so that the aeroplane is in a slight nose up atitude. The increase in pitch atitude helps to arrest the rate of descent and the reduction in thrust to zero reduces the speed. This procedure of reducing thrust and increasing pitch is known as the landing fare, although other terms like roundout are commonly used. The landing fare will allow the aeroplane to touch down onto the runway using the main wheels frst. It is important to understand that the technique of faring the aeroplane difers from one aeroplane to another and especially so between light general aviation aeroplanes and large commercial jet airliners.The second part of the landing is the ground run, ground roll or landing roll. This is the distance covered from touchdown until the aeroplane comes to full stop. As with the airborne section, there are a few critical actions that are carried out. Once the main wheels have setled onto the landing surface reverse thrust and lift spoilers can be activated and as the speed decreases further, the nose wheel will then setle onto the landing surface. Braking force is now applied and the aeroplane will slow to a stop. However, in normal operations, the aeroplane does not stop on the runway; rather the aeroplane is slowed to a safe speed where it can then be steered of the runway and taxied to the disembarkation point or ramp. The combined length 262Chapter 5 General Principles - Landingof the airborne section and the ground run or landing roll is known as the landing distance required. Pilots need to make sure that the landing distance required does not exceed the Landing Distance Available. LANDING DISTANCE AVAILABLE (LDA)The landing distance available is the distance from the point on the surface of the aerodrome above which the aeroplane can commence its landing, having regard to the obstructions in its approach path, to the nearest point in the direction of landing at which the surface of the aerodrome is incapable of bearing the weight of the aeroplane under normal operating conditions or at which there is an obstacle capable of afecting the safety of the aeroplane. Figure 5.2 Landing Distance AvailableIn short, the landing distance available is the length of runway from one threshold to another. These are not always at the edge of the runway, sometimes there are displaced thresholds which are some way in from the edge of the paved surface. You will recall that the landing distance starts at 50 ft. This point must be directly above the threshold. Landing on the threshold is not the aim of the landing.LIFT AND WEIGHTIn order to beter understand landing performance we need to analyze the forces acting upon the aeroplane, and how these forces might be modifed throughout the landing. The frst of the forces that we will consider is weight. As you have learnt already, weight acts vertically downwards towards the centre of the earth from the centre of gravity. During fight, weight is mainly balanced by lift. However, once the aeroplane is on the ground weight is balanced by the reaction of the ground acting up through the wheels on the undercarriage. The weight of the aeroplane on landing will be less than at take-of due to the fact that fuel has been consumed. There is however, a maximum structural landing mass which must not be exceeded. Lets now consider lift. Whilst lift helps to balance weight when in the air, once the aeroplane is on the ground, lift is no longer required. In fact during the landing roll, lift is detrimental to 263Chapter 5 General Principles - Landingthe landing performance. Producing lift will reduce the load placed on wheels and therefore decrease the braking efect. In large commercial aeroplanes once the main wheels have setled on the runway the lift spoilers or lift dumpers are deployed which act very quickly to disrupt the airfow over the wing and destroy lift.REVERSE THRUSTWhilst in level fight the forward acting force of thrust is essential to maintain sufcient speed for the wings to provide lift. During landing, because the aim is to bring the aeroplane to a stop, the thrust force must be reduced to zero. Any residual thrust would be detrimental to the landing performance. However, large propeller and jet aeroplane engines have the capability of redirecting the force of thrust in order to generate a braking efect on the aeroplane. This is known as reverse thrust. Reverse thrust helps to reduce the aeroplanes forward speed. Reverse thrust capability is especially important in conditions where braking force is reduced due to ice or water contamination on the runway.JET ENGINESJet engines produce reverse thrust by using one of several methods, but all jet engines follow the same basic principle which is to redirect the jet efux in a forwards direction. Many modern aeroplanes have a safety device whereby reverse thrust is not activated until a certain value of the aeroplanes weight is pressing down on the main wheels and until the wheels have reached a certain speed of rotation. Once the aeroplanes reverse thrust system has detected this, reverse thrust is activated and the engine will reconfgure itself so that the exhaust gas fow can be redirected forwards. You can see the extent of the re-confguration in Figure 5.3.Figure 5.3 An illustration of a jet engine during reverse thrust mode.However, pilots needs to recognize that the process of geting the engines reconfgured to generate full reverse thrust takes time, and so the aeroplane will have travelled a small distance from the touchdown point before reverse thrust takes efect. This fact reduces the efective time period during which reverse thrust can be used. Therefore, the efectiveness of reverse thrust on landing is reduced. 264Chapter 5 General Principles - LandingIt is important to note too that jet engine reverse thrust cannot be maintained right up to the point when the aeroplane comes to a full stop. Reverse thrust must de-activated before the forward speed reduces to a minimum value. As the aeroplane slows down the redirected airfow may start to be re-ingested into the compressor. This means that the jet engine recycles its own gas fow which signifcantly increases the engine temperatures but it also means that debris on the runway can be sucked into the engine, potentially causing major damage. As a result of this danger to the engines at low forward speeds reverse thrust be must deactivated below about 50 kt.The combined efect of late reverse thrust activation and early reverse thrust deactivation means that the time period for which reverse thrust can be used may be quite short.PROPELLER AEROPLANESLarge turbo propeller aeroplanes are also able to generate reverse thrust by redirecting the air fow forwards, but they do so in a very diferent way than that employed by a jet engine. For forward fight the propeller blade is angled in such a way as to displace air backwards thus producing forward thrust. In order for the propeller blade to direct air forwards and create a rearward acting force the blade angle must change. The blade angle required for the propeller to generate reverse thrust is known as reverse pitch. Because only a change of blade angle is required, a propeller aeroplane can switch from forward to reverse thrust far quicker than a jet aeroplane. This means that a propeller aeroplane can use reverse thrust a earlier in the landing roll than a jet aeroplane. A propeller aeroplane can also maintain reverse thrust until the aeroplane comes to a full stop. This capability gives the propeller aeroplane a greater braking advantage over the jet aeroplane during landing. In summary, the usable period of reverse thrust in the landing roll is shorter for a jet aeroplane than for a propeller aeroplane. For this reason, the authorities have laid down less stringent landing performance regulations for propeller aeroplanes. The precise nature of these regulations will be discussed later.DRAGLet us now consider the action of the drag force during landing. You may recall that there are several forms of drag. The main types of drag are parasite drag and induced drag. However, whilst the aeroplane is on the ground, during take-of and landing, wheel drag must be considered alongside the aerodynamic drag. The aim of the landing is to bring the aeroplane to a stop safely within the confnes of the runway. In order to decelerate, sufcient rearward directed forces need to act on the aeroplane. Consequently, in addition to reverse thrust aerodynamic drag plays a very crucial role in landing. INDUCED DRAGOf the two types of aerodynamic drag, well deal with induced drag frst. Induced drag is dependent on lift and is proportional to angle of atack. During the airborne section of the landing, there is still a large amount of lift being generated and the angle of atack is relatively high. This means that induced drag is far higher than in cruising fight. However, when the aeroplane nose wheel touches the runway, the angle of atack is almost nil. Induced drag is consequently reduced to almost zero. 265Chapter 5 General Principles - LandingPARASITE DRAGThe second form of aerodynamic drag is parasite drag. Parasite drag is a function of the aeroplanes forward facing cross sectional area, more accurately known as form drag, and of the aeroplanes forward speed. The confguration of the aeroplane for landing is such that the faps and slats are fully extended which signifcantly increases the aeroplanes form drag and therefore signifcant increases parasite drag. Once the aeroplane touches down, and the spoilers and speed brakes are deployed parasite drag increases even further. However, as the speed rapidly decays after touch down, so will the parasite drag; eventually decreasing to zero once the aeroplane reaches a full stop.In summary then, the aerodynamic drag comprising induced drag and parasite drag will be very high during the early part of the of the landing, but very soon after touch down will decay rapidly. WHEEL AND BRAKE DRAGHaving discussed aerodynamic drag, lets now consider wheel drag and brake drag. Wheel drag is the friction force between the wheel and the runway and with the wheel bearings, whereas brake drag is the friction force between the brake discs and the brake pads. Wheel drag will come into play as soon as the aeroplane touches down on the runway. However, as friction is a function of the force pushing two surfaces together, because there is still a lot of lift being generated, wheel load is small during the initial part of the landing run, and therefore wheel drag is also small. As speed reduces and as lift is destroyed by the spoilers, the wheel load increases which in turn increases the wheel drag. Therefore wheel drag increases throughout the landing roll and will reach a maximum value just before the aeroplane comes to rest.Brake drag is by far the most important and the most efective of the various drag forces during the landing since it provides the greatest retarding force. However, brake drag is only efective if there is also sufcient wheel drag or wheel friction between the tyres and the runway. If wheel drag is low, brake drag will also be low. Consequently, the brakes are efective only if there is sufcient friction between the tyres and the runway. During the early part of the landing run when there is not much load on the wheels and therefore not much wheel friction. Brake drag is consequently inefective in slowing down the aeroplane. However, as the lift reduces and more weight is placed on the wheels brake drag does become more efective in slowing down the aeroplane. Therefore, brake drag increases as the landing roll progresses. This concept explains why pilots need to destroy lift as soon as possible after touchdown so that the braking action can be at its peak efectiveness early on in the landing.In most large commercial aeroplanes however, the braking action may not actually be carried out by the pilots. Instead it can be carried out by a highly efective automatic anti skid braking system. This braking system can be set to low, medium or high braking levels, or levels 1 through to 3, and is especially important to use when landing on contaminated runways. 266Chapter 5 General Principles - LandingFigure 5.4 A graph showing the variations in drag through the landing rollIn summary then, the frst of the drag forces we discussed was aerodynamic drag. As speed decreases during the landing run, aerodynamic drag comprising of parasite and induced drag decreased. Brake drag on the other hand increased during the landing roll as more load was placed on the wheels. Notice that during the early part of the landing roll, aerodynamic drag provides the majority of the drag, whereas once the speed has dropped below 70% of the landing speed, brake drag provides the majority of the drag. The last line on the graph is total drag. From this line you can see that during the landing roll, total drag increases. Overall, you can see how important brake drag is. If the brakes were to fail, or the landing surface is very slippery, then the loss of braking would cause the landing performance to massively deteriorate therefore increasing the landing distance.LANDING DISTANCE FORMULANow well examine what determines the landing distance, and also what roles the four forces play. In order to do this, lets briefy detail the formula that is used to calculate the landing distance. Using Figure 5.5 you can see the landing distance formula. The leter s is the displacement or distance required to stop from a specifed speed which is V with a given deceleration d. Deceleration is force divided by mass. The force is aerodynamic drag, plus the braking co-efcient which is a function of wheel load, minus thrust, or in the case of reverse thrust, plus thrust. Figure 5.5 The expanded landing distance formula267Chapter 5 General Principles - LandingExpanding the formula in this manner allows you to see how a change in one of the variables has a knock on efect on the landing distance. Well now analyse in detail all the factors that can afect the landing, with our principle concern being how these factors afect the landing distance.EFFECT OF VARIABLE FACTORS ON LANDING DISTANCEWEIGHTThe mass of the aeroplane afects: the stalling speed and hence VREF

the deceleration for a given decelerating force the wheel dragIncreased mass increases stalling speed, and reduces the deceleration for a given decelerating force, both efects increasing the landing distance. Increased mass increases the brake drag available (if not torque limited) and this decreases the landing distance. The net efect is that the landing distance will increase with increasing mass, but to a lesser degree than the increase of take-of distance with increasing mass.DENSITYThe air density afects: the TAS for a given IAS the thrust or power of the engineAs thrust is small the main efect will be on the TAS. Low density (high temperature low pressure or high humidity) will give an increase in the landing distance due to the higher TAS, but again to a lesser degree than for the take-of distance.WINDYou will recall the headwinds decrease the true ground of the aeroplane for any given indicated airspeed. Thus, during a headwind, the forward speed over the landing surface is much less and as a result the distance required to bring the aeroplane to rest is decreased. A tailwind will have the opposite efect and it will increase the ground speed for a given indicated air speed. Thus, during a tailwind, the forward speed over the landing surface is much greater and as a result the distance required to bring the aeroplane to rest is increased. When examining the efects that winds have on landing performance, it is recommended that you do not use the actual wind that is given just in case the wind changes to a worse condition than the one you have planned for.When calculating actual landing distances, it is recommended that you assume only 50% of the headwind component, and 150% of the tailwind component. Most performance graphs that calculate the landing distance have the 50% headwind and 150% of recommendation already applied. An important note is that there is no allowance for crosswinds and therefore no safety factor for crosswinds. Crosswinds present an additional complication because of the efect of potentially crossed controls which would have to be applied because the wind will tend to push the aeroplane of the centreline. These issues mean that aeroplanes are given maximum crosswind limits.268Chapter 5 General Principles - LandingFLAP SETTINGFlaps are trailing edge devices used to increase the camber of the wing and generate more lift, thereby reducing the landing speed. However, the use of a lot of fap will dramatically increase aerodynamic drag as well. This is a beneft in landing and helps to slow the aeroplane down. Lets take a look at how diferent fap angles can afect the landing distance.With no fap the aircraft will have a much faster approach speed, and have very litle aerodynamic drag, therefore the landing distance will be large. However, with some fap selected, the approach speed is less, there will be more aerodynamic drag and consequently the landing distance will be smaller. However, with full fap, the approach speed is minimised and the aerodynamic drag is maximized. Therefore, the landing distance will be least compared to other fap setings. Having full fap selected for landing allows the aeroplane to maximize its landing mass. However, there is a disadvantage with having full fap if there was situation where the aeroplane needed to abort the landing and go-around for another atempt. You will recall that large fap angles greatly deteriorate the climb performance compared to no faps. Therefore in a go-around, after the climb has been established, retract the faps as soon as possible in the manner prescribed in the aeroplane fight manual.RUNWAY SLOPEThe next factor that can afect the landing performance is the slope of the runway. Slope has an impact on the landing distance because of its afect on how the component of weight acts along the longitudinal axis of the aeroplane. When the aircraft is on an upslope, the weight still acts towards the centre of the earth, but there is a component of weight, which acts in the direction of drag, that is, backwards along the longitudinal axis. This will increase the deceleration of the aeroplane and therefore decrease the landing distance. Conversely when the aeroplane is on a down slope, the weight component now acts in the direction of thrust. This adds to the forward force, therefore it will decrease the deceleration and increase the landing distance.A rough calculation to help quickly quantify the efect of slope is to assume that for every 1% slope, the landing distance is afected by 5%, or a factor of 1.05. The slope at some airfelds means that the landing may only be possible in one direction. These are called unidirectional runways. Treat these runways with extra caution and always check that the current wind velocity will still allow a safe landing since you may be forced to land with a tailwind.RUNWAY SURFACEMost landing performance graphs assume a paved hard surface. If the condition of the runway is not like this, then the efect on the landing distance needs to be understood and corrections applied.GRASSA lot of small airfelds have grass runways. The grass will increase the drag on the wheels. This is known as impingement drag and it will help to decelerate the aeroplane. However, grass severely reduces the wheel friction to the runway compared to a paved runway and therefore the wheel cannot be retarded efciently by the brakes, otherwise the wheel will lock and the wheel friction with the runway will reduce even further. 269Chapter 5 General Principles - LandingFigure 5.6 An illustration showing that for the wheel to advance grassmust be pushed out of the wayThe overall efect is that grass runways will increase the landing distance. In light general aviation aeroplanes, this increase in landing distance is about 15% compared to a landing on paved surface. However, most landings that you will carry out throughout your professional career will undoubtedly be on hard paved runways.CONTAMINATIONSIf the runway is covered, partially or fully by contaminants such as standing water, snow, slush or ice, then pay special atention to efect that they will have on the landing distance. These substances will have two main efects. The frst efect is that they will create impingement drag, much like grass did, but more importantly, the second efect is that these substances will substantially reduce the friction between the wheel and the runway. Therefore the wheel cannot be retarded efciently by the brakes. As a result of the reduced friction and therefore the subsequent reduced braking action, any contamination of the runway due to water, snow, slush or ice will signifcantly increase the landing distance. You can see the efect of the various contaminations to the landing distance in Figure 5.7270Chapter 5 General Principles - LandingFigure 5.7 A graph showing the efect of contamination on the landing distanceTypically on the a dry runway the braking co-efcient of friction is between 0.8 to 1.0, but on wet, slippery or icy runways, the braking co-efcient of friction can fall to less than 0.2. Because of the lack of efective braking on slippery surfaces, the aerodynamic drag and reverse thrust become more important in bringing the aeroplane to a stop as shown in the diagram to the right. On fooded or icy runways, reverse thrust accounts for 80% of the deceleration force.Figure 5.8 Various codes and terminology associated with runway contaminationsAn important point to note is that there is a distinction between the defnitions of a damp, wet or contaminated runway. A runway is considered to be damp when there is moisture present on the surface which changes its colour, but insufcient moisture to produce a refective surface.A wet runway is one whose moisture level makes the runway appear refective, but there are no 271Chapter 5 General Principles - Landingareas of standing water in excess of 3mm deep. Wet runways can cause the average dry landing distance to increase by as much as than 50%.A contaminated runway is one where more than 25% of the runway is covered in a layer of moisture, whose specifc gravity is equivalent to a depth of 3mm or more of water. The importance of runway contamination cannot be stressed enough, as many fatal accidents may have been avoided had due account been taken of the situation.From what you have learnt so far it is vital that the pilot be aware of the type of runway contamination, its depth, its extent, its efect on the braking and of course its overall efect on the operation concerned, in this case, the landing. The information on the runway contamination and braking efect can be given to the pilot through a report. These reports are either by snowtam, runway state code PIREPS or spoken by air trafc control and may include braking action or braking co-efcient. Use any runway reports together with your aeroplane fight manual or standard operating procedure to best gauge the landing technique and landing performance.HYDROPLANINGThere are three principal types of aquaplaning or hydroplaning as it is now more commonly known. DYNAMIC HYDROPLANINGThe frst is Dynamic hydroplaning. When an aircraft lands fast enough on a wet runway with at least 1/10 inch of standing water, inertial efects prevent water escaping from the footprint area, and the tyre is buoyed or held of the pavement by hydrodynamic force. Most people have experienced this type of hydroplaning when they have driven over a patch of water at high speed. The speed at which dynamic hydroplaning occurs is called VP. There is simple formula to help calculate the dynamic hydroplaning speed. For rotating tires the dynamic hydroplaning speed in knots is equal to 9 times the square root of the tyre pressure in PSI. For a typical 737 the dynamic hydroplaning speed is between 90 and 120 knots.However, for non rotating tires the dynamic hydroplaning speed is equal to 7.7 times the square root of the tyre pressure. The danger from hydroplaning is the virtually nil braking and steering efect. The most positive methods of preventing this type of hydroplaning is to groove the tires, transversely groove the runway, ensure the runway pavement is convex from the centre line and ensure the runway has a macro-texture.VISCOUS HYDROPLANINGThe second type of hydroplaning is called viscous hydroplaning. It occurs because of the viscous properties of water acting like a lubricant. A thin flm of fuid not more than 1/1000 of an inch deep cannot be penetrated by the tire in the footprint area and the tire rolls on top of the flm. Viscous hydroplaning can occur at a much lower speed than dynamic hydroplaning, but it requires a smooth surface. The most positive method of preventing this type of hydroplaning is to provide a micro-texture to the pavement surface which breaks up the flm of water allowing it to collect into very small pockets. This means that the tyre footprint will sit on the peaks of the textured surface and not the flm of water. 272Chapter 5 General Principles - LandingREVERTED RUBBER HYDROPLANINGReverted rubber hydroplaning is a complex phenomenon which over the years has been the subject of a variety of explanations. Reverted rubber hydroplaning requires a prolonged, locked wheel skid, reverted rubber, and a wet runway surface. The locked wheels create enough heat to vaporize the underlying water flm forming a cushion of steam that lifts the tire of the runway and eliminates tyre to surface contact. The steam heat reverts the rubber to a black gummy deposit on the runway. Once started, reverted rubber skidding will persist down to very low speeds, virtually until the aircraft comes to rest. During the skid there is no steering capability and the braking efect is almost nil. Reverted rubber hydroplaning is greatly reduced in modern aeroplanes due to the standardization of advanced anti skid braking systems which prevent wheel lock up.LANDING TECHNIQUE ON SLIPPERY RUNWAYSDetailed below is some advice and guidance to how to land on contaminated runways. Firstly, check the current weather, and the runway conditions using the most accurate information possible. Once this has been done, completely re-asses the landing performance data to ensure satisfactory compliance to the regulations. Ensure you are at VREF at the landing screen height and prepare to land the aircraft in the touchdown zone within the 1,000 ft target of the airborne segment. Land on the centre line with minimal lateral drift and without excess speed.Arm auto spoilers and auto brakes as appropriate which assure prompt stopping efort after touchdown. For aeroplanes fted with automatic anti skid brakes, the brakes will be applied above the dynamic hydroplaning speed, but for aeroplanes without this system, only apply the brakes below the dynamic hydroplaning speed. The fare should lead to a frm touchdown, sometimes described as fying the aeroplane onto the runway. Positive landings will help place load on the wheels which will increase braking efectiveness and squeeze out the water from the tire foot print area. Do not allow the airplane to foat and do not atempt achieve a perfectly smooth touchdown. An extended fare will extend the touchdown point. Soft touchdowns will delay wheel spin up and delay oleo compression which is needed for auto brake and auto spoiler activation.After main gear touchdown, do not hold the nose wheel of the runway. Smoothly fy the nose wheel onto the runway by relaxing aft control column pressure. Deploy spoilers as soon as possible after touchdown or confrm auto spoiler deployment. If the aeroplane does not have auto brake then initiate braking once spoilers have been raised and nose wheel has contacted the runway. Apply brakes smoothly and symmetrically.Initiate reverse thrust as soon as possible after touchdown of the main wheels and target the rollout to stop well short of the end of the runway. Leave a margin for unexpectedly low friction due to wet rubber deposits or hydroplaning.273Chapter 5 General Principles - LandingMICROBURSTS AND WINDSHEAROf all the phases of fight it is the landing phase which is the most susceptible to severe weather conditions. The low altitude, low speed, low thrust setings and high drag during the landing phase mean that the aeroplane is at its most vulnerable. Landing when windshear is observed or forecast should be avoided, but if conditions are within limits, then landing at a higher speed should be used, but caution should be exercised since the higher landing speed will greatly increase the distance of the airborne section and ground roll of the landing.Figure 5.9 An illustration of a microburst at an airfeldThe microbust is the single most hazardous weather phenomenon to aeroplanes close the ground. We briefy discussed the microburst phenomenon in the take-of section, but in landing it represents an even greater threat. However, there are certain clues which can give a pilot advance warning. The most obvious clue to a potential microbust are the tendrils of rain found beneath a storm cloud which are called virga or fallstreaks. Virga is precipitation falling but evaporating before it reaches the ground as shown here.In the absence of specifc guidance, here are some suggested techniques for identifying and dealing with a microburst encounter whilst on the approach to land. Using Figure 5.10 you can initially see that there will be an increase in airspeed and a rise above the approach path caused by the increasing headwind. On modern aeroplanes a windshear alert is usually given. This should be seen as the precursor to the microburst. Any hope of stabilized approach should be abandoned and a missed approach should be initiated. Without hesitation the power should be increased to go around power, the nose raised and the aeroplane fown in accordance with the missed approach procedure. Typically the nose atitude will be about 15 and the control column should be held against the bufet or stick shaker. The initial bonus of increased airspeed may now be rapidly eroded as the downdraught is encountered. Airspeed will fall and the aeroplane may start to descend despite high power and high pitch angle. 274Chapter 5 General Principles - LandingFigure 5.10 An illustration showing the fight path of a typical aeroplane on theapproach to land during an microburst encounterThe point at which the tailwind starts to be encountered may be the most critical. The rate of descent may reduce, but the airspeed may continue to fall and any height loss may now break into obstacle clearance margins. Maximum thrust is now usually needed and the nose atitude kept high on the stall warning margins to escape the efects of the tailwind. Had the go-around not been initiated at the early warning stage, then it is highly unlikely that the aeroplane with survive the encounter. Because windshear, heavy rain, poor visibility, runway contamination and microbusts are hazards very closely associated with thunderstorms, it is advisable never to land with a thunderstorm at or in the immediate vicinity of the aerodrome. Delay the landing or consider diverting. There are too many cases to list of pilots who have atempted to land in bad weather and they have sadly perished along with many passengers. Most of these occurrences were avoidable had due account been made of the situation. 275Chapter 5 General Principles - LandingQUESTIONS1. The landing reference speed VREF has, in accordance with international requirements, the following margins above the stall speed in landing confguration for a Class B aeroplane:a. 15%b. 20%c. 10%d. 30%2. At a given mass, the stalling speed of a twin engine, class B aeroplane is 100 kt in the landing confguration. The minimum speed a pilot must maintain in short fnal is:a. 130 ktb. 115 ktc. 125 ktd. 120 kt3. The stalling speed or the minimum steady fight speed at which the aeroplane is controllable in landing confguration is abbreviated as:a. VS1b. VSc. VMCd. VSO4. What margin above the stall speed is provided by the landing reference speed VREF for a Class B aeroplane?a. 1.10 VSOb. VMCA x 1.2c. 1.30 VSOd. 1.05 VSO5. An increase in atmospheric pressure has, among other things, the following consequences on landing performance:

a. a reduced landing distance and degraded go around performance.b. a reduced landing distance and improved go-around performance.c. an increased landing distance and degraded go-around performance.d. an increased landing distance and improved go-around performance.276Chapter 5General Principles - Landing6. The take-of distance of an aircraft is 600 m in a standard atmosphere, with no wind and at a pressure altitude of 0 ft. Using the following corrections: 20 m / 1,000 ft feld elevation- 5 m / kt headwind+ 10 m / kt tail wind 15 m / % runway slope 5 m / C deviation from standard temperatureThe take-of distance from an airport at 1,000 ft elevation, temperature 17C, QNH 1013.25 hPa, 1% up-slope, 10 kt tailwind is?a. 555 mb. 685 mc. 755 md. 715 m7. To minimize the risk of hydroplaning during the landing the pilot of a modern airliner should:a. make a positive landing and apply maximum reverse thrust and brakes as quickly as possible.b. use maximum reverse thrust, and should start braking below the hydroplaning speed.c. use normal landing, braking and reverse thrust techniques.d. postpone the landing until the risk of hydroplaning no longer exists.8. An aircraft has two certifed landing faps positions, 25 and 35. If a pilot chooses 25 instead of 35, the aircraft will have:a. a reduced landing distance and beter go-around performance.b. an increased landing distance and degraded go-around performance.c. a reduced landing distance and degraded go-around performance.d. an increased landing distance and beter go-around performance.9. May anti-skid be considered to determine the take-of and landing data?a. Only for take-of.b. Only for landing.c. Yes.d. No.10. An aircraft has two certifed landing faps positions, 25 and 35. If a pilot chooses 35 instead of 25, the aircraft will have:a. an increased landing distance and beter go-around performance.b. a reduced landing distance and degraded go-around performance.c. a reduced landing distance and beter go-around performance.d. an increased landing distance and degraded go-around performance.277Chapter 5 General Principles - Landing278Chapter 5General Principles - LandingANSWERS1. D2. A3. D4. C5. B6. C7. A8. D9. C10. B279Chapter 6Single Engine Class B AircraftTake OfCHAPTER SIXSINGLE ENGINE CLASS B AIRCRAFT - TAKE OFFContentsPERFORMANCE CLASS B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281GENERAL REQUIREMENTS (JAR - OPS 1.525) . . . . . . . . . . . . . . . . . . . . . 281TAKE-OFF DISTANCE (JAR 23.53 ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281TAKE-OFF REQUIREMENTS/ FIELD LENGTH REQUIREMENTS . . . . . . . . . . 282FACTORS TO BE ACCOUNTED FOR. . . . . . . . . . . . . . . . . . . . . . . . . . . 283SURFACE CONDITION FACTORS . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284PRESENTATION OF DATA. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285CERTIFICATION REQUIREMENTS. . . . . . . . . . . . . . . . . . . . . . . . . . . . 288QUESTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289ANSWERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292280Chapter 6Single Engine Class B AircraftTake Of281Chapter 6Single Engine Class B AircraftTake OfPERFORMANCE CLASS BJust to remind yourself, Class B single engine aeroplanes are propeller driven aeroplanes with a maximum approved passenger seating confguration of 9 or less, and a maximum take-of mass of 5,700 kg or less.JAR 23 contains requirements for normal, utility, and aerobatic category aeroplanes, and also for commuter category aeroplanes.A lot of the detail about the regulations for the various classes of aeroplanes can be found in CAP 698 which should have been issued to you by your training provider. Make sure you are aware of what CAP 698 gives you. Those bits of information that are included in this book and that are also found in CAP698 will be mentioned to make it easier for you to identify those areas which do not need to be commited to memory.GENERAL REQUIREMENTS (JAR - OPS 1.525)These requirements can be found in CAP 698 on page 1 of section 2 under paragraph 1.2. There are four general requirements about operating Class B single engine aeroplanes for commercial air transport purposes. The frst is that this aeroplane shall not be operated at night. The second that the aeroplane must not be operated in Instrument Meteorological Conditions (IMC) except under special visual fight rules (SVFR). The third is that it must not be operated unless suitable surfaces are available en-route which permit a safe forced landing to be made should the engine failure occur at any point on the route. Lastly, that this type of aeroplane must not be operated above a cloud layer that extends below the relevant minimum safe altitude. The reason why the later regulation exists is quite easy to understand. If the engine were to fail during these conditions it would be almost impossible for a pilot to be able to see the landing surface and therefore impossible for the pilot to carry of a safe forced landing.TAKE-OFF DISTANCE (JAR 23.53 )The gross take-of distance for Class B aeroplanes (other than those in the commuter category) is the distance from the start of take-of to a screen height of 50 ft above the take-of surface, with take-of power set, rotating at VR and achieving the specifed speed at the screen.The rotation speed VR, must not be less than VS1The take-of safety speed (screen height speed) must be not less than the greater of: a speed that is safe under all reasonably expected conditionsor 1.2 VS1282Chapter 6Single Engine Class B AircraftTake OfTAKE-OFF REQUIREMENTS/ FIELD LENGTH REQUIREMENTSThere is only one take-of requirement for single engine Class B aeroplanes. The requirement is that the mass of the aeroplane must be such that the take-of can be completed within the available distances. In other word the take-of must be complete within the feld length available. This requirement is called the Field Length Requirement. The Field Length Requirements are detailed below and you can fnd them in CAP 698 under paragraph 2.1.1 of page 1 and 2 of section 2. When no stopway or clearway is available, the take-of distance when multiplied by 1.25 must not exceed TORA (Gross TOD x 1.25 must not exceed the TORA) When a stopway and/or clearway is available the take-of distance must: not exceed TORA (Gross TOD must not exceed the TORA) when multiplied by 1.3, not exceed ASDA (Gross TOD x 1.3 must not exceed the ASDA) when multiplied by 1.15, not exceed TODA (Gross TOD x 1.15 must not exceed the TODA)To understand these requirements it might be easier to work through and example. EXAMPLE: Let us assume that the aeroplane fight manual gives the take-of distance as 3000ft and that there is no stopway or clearway available at the airport. What is the net take-of distance or the minimum length or TORA?In this case we must satisfy the requirement that when no stopway or clearway is available, the take-of distance when multiplied by 1.25 must not exceed TORA (Gross TOD x 1.25 must not exceed the TORA)To carry out the calculation, multiply 3000 ft by 1.25 gives us a value of 3750 ft. Essentially this means that the runway must be at least 3750 ft long, or, more correctly, the take of run available (TORA) must be at least 3750 ft long. 283Chapter 6Single Engine Class B AircraftTake OfFigure 6.1 The Gross take-of distance multiplied by 1.25 must not exceed the TORAThis rule ensures that once the aeroplane has completed the take-of, that is, at the screen height, there should be at least 25% of the runway remaining. Therefore the regulation requirement provides an adequate safety margin for the take-of. We discussed these safety concerns during the lesson on performance standards. To interpret the regulation, although the actual take-of distance is 3000 ft, the authorities are suggesting the worse case in a million would be a take-of that is 3750 ft long. This helps us again to see the diference between net and gross performance. Here the gross performance for the take-of is 3000 ft but the net performance, which is always worse performance, is 3750 ft.FACTORS TO BE ACCOUNTED FORThe regulations in CS-23 state that when calculating the gross take-of distance, in other words, before we add the factors mention previously, certain details must be accounted for. These are listed below. The gross take-of distance required shall take account of: the mass of the aeroplane at the start of the take-of run the pressure altitude at the aerodrome the ambient temperature at the aerodrome the runway surface conditions and the type of runway surface the runway slope not more than 50% of the reported headwind component or not less than 150% of the reported tailwind component 284Chapter 6Single Engine Class B AircraftTake OfSURFACE CONDITION FACTORSConcentrating on point d) and e) from above the regulations stipulate that due account must be made of the runway condition. Most performance data in the aeroplane fight manual assumes a level, dry and hard runway. Therefore, correction factors must be applied to the gross take-of distance when the runway conditions are diferent. There are various correction factors such as for grass runways, wet runways and runways which are sloped. These factors are detailed next.At the top of page 2 section 2 under section c) in CAP 698 it states that if the runway is other than dry and paved the following correction factors must be used when determining the take-of distance. These are shown for you in fgure 6.2Figure 6.2 The corrections factors that need to be applied to the gross take-of distancewhen the runway is other than dry, paved and level.As you have learnt already, grass runways will increase the take-of distance compared to paved runways. Here the factor to use to account for the efect of dry grass is 1.2 and 1.3 if the grass is wet.At the top of page 2 section 2 underneath the table, point d details the corrections to be applied if there is a slope to the runway. It states that a pilot must increase the take-of distance by 5%, or by a factor of 1.05, for ever 1% upslope. However, it also states that no factorisation is permited for downslope. In other words, when an aeroplane may be taking-of on a downwards sloping runway no correction factor is to be applied for the downslope. The reason for ignoring the downslope is because a downslope will decrease the take-of distance. This helps to add a litle extra safety to the take-of distance calculation.285Chapter 6Single Engine Class B AircraftTake OfPRESENTATION OF DATAThe take-of distance required is usually presented in graphical form. You can see such a graph by looking at fgure 2.1 on page 3 of section 2 in CAP 698. Firstly, the title of the graph indicates that the graph is for calculating the gross take-of distance with no faps selected. For other fap setings you may have to consult another graph. Take a look at the associated conditions paying particular notice to the power and fap setings as shown here. Also notice that this graph assumes a runway that is paved, level and dry. If the runway conditions required for a given calculation are diferent, then corrections will need to be made to the values this graph will give. We briefy saw these correction factors earlier. Notice a small box in the middle of the graph which highlights the rotation and screen height speeds for diferent weights. A pilot must adhere to these accurately because they are the speeds which have been used to construct this graph. If a pilot were to deviate from these speeds, the required aircraft performance would not be achieved.286Chapter 6Single Engine Class B AircraftTake OfFigure 6.3 An example take-of distance calculation287Chapter 6Single Engine Class B AircraftTake OfHaving looked at the various bits of information around the graphs lets now examine the graph itself. The left hand carpet of the graph involves the variations of temperature and pressure altitude. This part of the graph accounts for the efect of air density on the take-of distance. The middle carpet accounts for the efect of the mass and to the right of this carpet is the wind correction carpet. Notice the diferences in the slope of the headwind and tailwind lines. This means the 150% and 50% wind rules have been applied. The last carpet on the far right of the graph is labelled obstacle height. Although there is no obstacle as such at the end of the take-of run, you will recall that the take-of is not complete until a screen height of 50 ft is reached.Using an example we will work through the graph so that you are able to see how the take-of distance is calculated. For this example, follow through the red line in fgure 6.3. In the example we will assume a temperature of 15C at a given airfeld 4000ft above mean sea level. To use the graph then, move upwards from 15C until you have reached the 4000ft pressure altitude line. Then move right until you meet the frst reference line. The next variable is mass; in our example let us assume a mass of 3400 lbs. From the reference line we must move down along the sloping guidelines until we reach 3400 lbs as shown here.From this point we go horizontally to the right until the next reference line. The next variable is wind.In our example we will assume a 10 knot headwind. Notice though that the slopes of the headwind and tailwind lines are diferent. We therefore recognise that the 150% tailwind and 50% headwind rules that you learnt about in the general principles for take-of lesson have already been applied. Travel down the headwind line until you reach 10 knots headwind and then move horizontally right once more, up to the last reference line. Remember that the take of is not complete until the aeroplane has reached the 50ft screen height, so from this point on the graph we must more up the guidelines to the end. The take of distance in our example, then, is approximately 2300ft from brake release to the 50ft screen.However, let us now assume that the runway has a grass surface and that the grass is wet. Remember that both these variables cause an increase in the take of distance. Since the graph assumes a paved dry runway we will have to make a correction to the take of distance for the diferent conditions. You may recall that wet grass increases the distance by 30% or by a factor of 1.3. Our calculated take of distance was 2300ft, but correcting it for wet grass would now make the take of distance 2990ft. Let us now assume the runway has an upslope of 1%. You may recall this would increase the take of distance by 5% or by a factor of 1.05.This now makes the total gross take of distance 3140ft. To comply with the feld length requirements and to obtain a net take-of distance, you may recall that the take-of distance must be compared to the available distances at the airfeld to ensure it does not exceed the limits laid down by the authorities. To remind you of these regulations turn to page 1 section 2 in the CAP 698.288Chapter 6Single Engine Class B AircraftTake OfCERTIFICATION REQUIREMENTSThe last part of the lesson will focus on some of the certifcation specifcations for single engine class B aeroplanes. These can be found in the document from EASA called CS 23. There are only two main specifcations that apply to the take-of phase of fight and these specifcations concern the take-of speeds.The frst of these certifcation specifcations concerns VR. You may recall that VR is the speed at which the pilot makes a control input with the intention of geting the aeroplane out of contact with the runway. The certifcation specifcations state that for the single engine aeroplane, the speed VR must not be less than VS1. VS1 being the stall speed or the minimum steady fight speed of the aeroplane obtained in a specifed confguration. The confguration concerned is that confguration used for the take-of.The second of the certifcation specifcation concerns the speed of the aeroplane at the screen height. The specifcations state that the speed at 15m or 50ft above the take-of surface must be more than the higher of a speed that is safe under all reasonably expected conditions, or 1.2 VS1. You will recall from an earlier lesson that the speed that must be atained at the screen height is commonly referred to as the take-of safety speed.The certifcation regulations about VR and the take-of safety speed are not found in the CAP 698 and therefore must be commited to memory.289Chapter 6Single Engine Class B AircraftTake OfQUESTIONS1. For a single engine Class B aeroplane, how does runway slope afect allowable take-of mass, assuming other factors remain constant and not limiting?a. An uphill slope decreases take-of mass.b. Allowable take-of mass is not afected by runway slope.c. A downhill slope decreases allowable take-of mass.d. A downhill slope increases allowable take-of mass.2. For this question use Performance Manual CAP 698 SEP 1 Figure 2.1.With regard to the take-of performance chart for the single engine aeroplane, determine the take-of speed for (1) rotation and (2) at a height of 50 ft.Given:-O.A.T : ISA + 10Pressure Altitude: 5,000 ftAeroplane mass: 3,400 lbsHeadwind component: 5 ktFlaps: up Runway: Tarred and Drya. 73 and 84 ktb. 68 and 78 ktc. 65 and 75 ktd. 71 and 82 kt3. For this question use Performance Manual CAP 698 SEP 1 Figure 2.2.With regard to the take-of performance chart for the single engine aeroplane determine the take-of distance over a 50 ft obstacle height.Given:- O.A.T : 30C Pressure Altitude: 1,000 ft Aeroplane Mass: 2,950 lbs Tailwind component: 5 kt Flaps: Approach seting Runway: Short, wet grass, frm subsoilCorrection factor: 1.25 for the current runway conditionsa. 1,700 ftb. 2,500 ftc. 2,200 ftd. 1,900 ft290Chapter 6Single Engine Class B AircraftTake Of4. For this question use Performance Manual CAP 698 SEP 1 Figure 2.1.With regard to the take of performance chart for the single engine aeroplane determine the maximum allowable take of mass.Given:-O.A.T : ISAPressure Altitude: 4,000 ftHeadwind component: 5 ktFlaps: up Runway: Tarred and DryFactored runway length: 2,000 ftObstacle height: 50 fta. 3,000 lbsb. 2,900 lbsc. 3,650 lbsd. 3,200 lbs5. For this question use Performance Manual CAP 698 SEP 1 Figure 2.1.With regard to the take-of performance chart for the single engine aeroplane determine the take-of distance to a height of 50 ft.Given:- O.A.T : 30CPressure Altitude: 1,000 ftAeroplane Mass: 3,450 lbsTailwind component: 2.5 ktFlaps: up Runway: Tarred and Drya. approximately : 2,200 feetb. approximately : 2,400 feetc. approximately : 1,400 feetd. approximately : 2,800 feet6. For this question use Performance Manual CAP 698 SEP 1 Figure 2.2.With regard to the take-of performance chart for the single engine aeroplane determine the take-of distance to a height of 50 ft.Given:-O.A.T : -7CPressure Altitude: 7,000 ftAeroplane Mass: 2,950 lbsHeadwind component: 5 ktFlaps: Approach setingRunway: Tarred and Drya. approximately : 1,150 ftb. approximately : 2,450 ftc. approximately : 1,500 ftd. approximately : 2,100 ft291Chapter 6Single Engine Class B AircraftTake Of7. For this question use Performance Manual CAP 698 SEP 1 Figure 2.1.With regard to the take-of performance chart for the single engine aeroplane determine the take-of distance to a height of 50 ft.Given:-Airport characteristics: hard, dry and zero slope runway Pressure altitude: 1,500 ft Outside air temperature: +18C wind component: 4 knots tailwind Take-of mass : 1,270 kg

a. 520 mb. 415 mc. 440 md. 615 m8. For this question use Performance Manual CAP 698 SEP 1 Figure 2.2.With regard to the take-of performance chart for the single engine aeroplane determine the take-of distance to a height of 50 ft.Given:- O.A.T : 38CPressure Altitude: 4,000 ftAeroplane Mass: 3,400 lbsTailwind component: 5 ktFlaps: Approach setingRunway: Dry GrassCorrection factor: 1.2a. approximately : 3,250 ftb. approximately : 4,200 ftc. approximately : 5,040 ftd. approximately : 3,900 ft9. For a Class B aircraft at an aerodrome with no stopway or clearway, the minimum length of Take-of Run that must be available to satisfy the take-of requirements:a. Must not be less than the gross take of distance to 50ft.b. Must not be less than 1.15 times the gross take of distance to 50ft.c. Must not be less than 1.25 times the gross take of distance to 50ft.d. Must not be less than 1.3 times the gross take of distance to 50ft.10. For a single engine Class B aircraft, the rotation speed VR:a. Must not be less than 1.1 VS1b. Must not be less than VS1c. Must not be less than 1.2 VMCd. Must not be less than VMC11. For a single engine Class B aircraft at an aerodrome with stopway:a. The TOD x 1.3 must not exceed the ASDA.b. The TOD must not exceed the ASDA x 1.3.c. The TOD x 1.25 must not exceed the ASDA.d. The TOD must not exceed the ASDA x 1.25.292Chapter 6Single Engine Class B AircraftTake OfANSWERS1. A2. D3. C4. D5. B6. D7. A8. D9. C10. B11. A293Chapter 7Single Engine Class BClimbCHAPTER SEVENSINGLE ENGINE CLASS B - CLIMBContentsCLIMB PERFORMANCE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295PRESENTATION OF DATA. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295USE OF THE CLIMB GRAPH DATA . . . . . . . . . . . . . . . . . . . . . . . . . . . 296EXAMPLES IN CAP 698. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296QUESTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297ANSWERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298294Chapter 7Single Engine Class BClimb295Chapter 7Single Engine Class BClimbCLIMB PERFORMANCEIn this lesson we will concentrate on the regulatory requirements for the class B single engine aeroplane in the climb. These regulations are only applicable to operating the single engine aeroplane in the commercial air transport category. The take-of climb requirements for class b single engine aeroplanes are stated in CS-OPS 1, but, again, as for the take-of requirements, the climb regulations can be found in the CAP 698, on page 6 of section 2. At the top of the page 6 of section 2 under point 3.1 we read that there are no obstacle clearance limits, or minimum acceptable climb gradients. Let us break down the two elements in this statement.Firstly, the regulations are saying that there is no requirement for a single engine class B aeroplane to demonstrate that it can clear an obstacle within the take-of fight path. However, other classes of aeroplane have to demonstrate that obstacles within the take-of fight path can be cleared by a set limit of 50 feet or 35 feet. It seems strange, then, that there is no regulatory requirement for obstacle clearance in the case of a single engine Class B aeroplane. But, surely the idea of the regulations is to enforce a safety margin. What must be understood here, though, is that the pilot of a Class B single engine aeroplane must, at all times, have visual contact with the ground. Consequently the pilot will at all times be able to identify the obstacle within the take-of path and, therefore, avoid it. However, you may judge that it would still be prudent to carry out an obstacle clearance calculation if there were known obstacle in the take-of fight path.The second part of the regulation states that there is no minimum acceptable climb gradient for a single engine class B aeroplane. Again, this seems strange; the idea of the operational regulations is to enforce a safety margin. However, the reason why there is no operational regulation on the climb performance is that the certifcation specifcations for this type of aeroplane, as stated in CS-23, are stringent enough on their own.Even though there is no operational requirement for a minimum climb performance, it would not be safe to operate the aeroplane in such a manner that its performance is so poor it is barely able to climb. Therefore, it is important as a pilot operating this class and type of aeroplane to know what climb performance would be achieved so that the aeroplane will at least be able to climb sufciently. PRESENTATION OF DATAProvided in most pilot operating manuals or aeroplane fight manuals are climb graphs that help the pilot to calculate the gradient of climb. Such a graph can be found in CAP 698 in section 2 page 7 fgure 2.3. As with all graphs, before you use it, ensure you are familiar with the associated conditions of that graph. In this graph the throtles are at maximum, mixture is rich, faps and gear are up and the cowl faps set as required. The climb speed for the graph is taken to be 100 knots indicated airspeed for all masses.The graph has an example that you can follow using the dashed black lines so that you can practice on your own. Graph accuracy is very important so try and be as careful as possible when using them. Be especially careful when working out your true air speed which is need when you approach the right hand side of the graph. You may need your navigation computer to do this as it might not be given to you, such as the case in the example. 296Chapter 7Single Engine Class BClimbUSE OF THE CLIMB GRAPH DATAThere will be two main uses of the climb graph. The frst main use of the graph is to calculate the time to climb to the cruise altitude. For this, the pilot would need to know the cruise altitude and the rate of climb. However, if the gradient from this graph is used for the calculation of obstacle clearance or the ground distance, then the gradient must be adjusted for the efect of wind. This particular climb graph makes no correction for wind. The reason why the gradient must be corrected for wind is because obstacle clearance calculations or ground distance calculations use ground gradients, that is gradients measured relative to the ground and ground gradients are afected by wind.If you need to refresh your memory on the diference between ground gradients and air gradients and the efects of wind, then go back to the general performance principles climb chapter. EXAMPLES IN CAP 698There are some example calculations in CAP 698 for you to work through and practise. For example, on page 6 section 2 there is an example on how to calculate the climb gradient using the graph. Just below this, there is another example to determine what the maximum permissible mass is in order to achieve a 4% climb gradient. This maximum permissible mass is sometimes referred to as the MAT or WAT limit.On page 8 of section 2 is another example. This is an example of how to calculate the horizontal ground distance required to climb to a given height.297Chapter 7Single Engine Class BClimbQUESTIONS1. For this question use Performance Manual CAP 698 SEP 1 Figure 2.3.Using the climb performance chart, for the single engine aeroplane, determine the rate of climb and the gradient of climb in the following conditions:Given:- O.A.T at Take-of: ISAAirport pressure altitude: 3,000 ftAeroplane mass: 3,450 lbsSpeed: 100 KIASa. 1,310 ft/min and 11.3%b. 1,130 ft/min and 10.6%c. 1,030 ft/min and 8.4%d. 1,140 ft/min and 11.1%2. For this question use Performance Manual CAP 698 SEP 1 Figure 2.3.Using the climb performance chart, for the single engine aeroplane, determine the ground distance to reach a height of 1,500 ft in the following conditions:Given:-O.A.T at Take-of: ISAAirport pressure altitude: 5,000 ftAeroplane mass: 3,300 lbsSpeed: 100 KIASWind component: 5 kts Tailwinda. 19,250 ftb. 14,275 ftc. 14,925 ftd. 15,625 ft3. For this question use Performance Manual CAP 698 SEP 1 Figure 2.3.With regard to the climb performance chart for the single engine aeroplane determine the climb speed (ft/min. .Given:- O.A.T : ISA + 15CPressure Altitude: 0 ftAeroplane Mass: 3,400 lbsFlaps: upSpeed: 100 KIASa. 1150 ft/minb. 1290 ft/minc. 1370 ft/mind. 1210 ft/min298Chapter 7Single Engine Class BClimb4. For this question use Performance Manual CAP 698 SEP 1 Figure 2.3.Using the climb performance chart, for the single engine aeroplane, determine the ground distance to reach a height of 2,000 ft in the following conditions:Given:- O.A.T. at take-of: 25CAirport pressure altitude: 1,000 ftAeroplane mass: 3,600 lbsSpeed: 100 KIASWind component: 15 kts Headwinda. 15,290 ftb. 18,750 ftc. 16,410 ftd. 16,050 ftANSWERS1. B 2. D 3. B 4. D299Chapter 8Single Engine Class BEn Route and DescentCHAPTER EIGHTSINGLE ENGINE CLASS B - EN-ROUTE AND DESCENTContentsEN-ROUTE SECTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301EN-ROUTE AND DESCENT REQUIREMENTS . . . . . . . . . . . . . . . . . . . . . 301INFORMATION IN CAP 698 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305TYPICAL RANGE AND ENDURANCE GRAPHS. . . . . . . . . . . . . . . . . . . . 306QUESTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308ANSWERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310300Chapter 8Single Engine Class BEn Route and Descent301Chapter 8Single Engine Class BEn Route and DescentEN-ROUTE SECTIONThis chapter will focus on the Class B single engine performance requirements for the en-route and descent stage of fight.Essentially the en-route part of the fight is considered to be from 1500 ft above the airfeld from which the aeroplane has taken of to 1000 ft above the destination airfeld. Although the CAP 698 manual does show the en-route performance requirements these requirements are scatered through the document, which does not make for ease of reference.EN-ROUTE AND DESCENT REQUIREMENTSJAR-OPS 1 states that an operator must ensure that the aeroplane, in the meteorological conditions expected for the fight, and in the event of engine failure, is capable of reaching a place at which a safe forced landing can be made. In order to be able to comply with the rule, an operator has to know certain details about the route to be fown and the performance of the aeroplane. First of all, of course, for any given route, an operator must know the safe forced landing areas. The next detail that the operator needs to know is whether his aeroplane will be able to reach these areas if the engine were to fail while en-route. Whether or not this is possible will depend on two things; the altitude chosen for the fight the descent gradient of the aeroplane following engine failure. If these two parameters are known, it is possible to calculate how far the aeroplane will travel following engine failure. So, let us work through an actual example.Let us assume a cruise altitude of 10,000 feet and a gradient of descent of 7% following an engine failure. What is the descent range?We can work out the horizontal distance travelled or descent range by taking the height of the aeroplane above the ground dividing this by the gradient and then multiplying by 100.Horizontal Distance = Vertical/Gradient X 100Figure 8.1 An illustration showing an example descent range calculation302Chapter 8Single Engine Class BEn Route and DescentIn this case we see that in still air the aeroplane will cover in the glide a horizontal distance of 142,857ft or roughly 23.5 nm following engine failure. This means that the aeroplane must not pass further away than 23.5 nm from any safe forced landing locations. So then draw a circle of 23.5 nm radius, around each of the safe forced landing locations (the yellow dots) as you can see in fgure 8.2. Then draw a track line from Airfeld A to Airfeld B that is within each circle. If the fight track falls outside of the circles then, following engine failure the aeroplane will not make it to a safe forced landing area.Figure 8.2 An illustration showing the glide descent range around each forced landing area and the subsequent track that is required to remain with glide range of each feld.If the aeroplane were to operate at a higher altitude, then it would be able to cover a greater distance in the glide following an engine failure. For example, operating at 15,000 ft instead of 10,000ft would increase the aeroplanes still air glide range to 35 nautical miles. Therefore the circles around each forced landing location will grow as shown in fgure 8.3.303Chapter 8Single Engine Class BEn Route and DescentFigure 8.3 An illustration showing the glide descent range around each forced landing area and the subsequent track that is required to remain with glide range of each feld.Notice that now that the aeroplane can fy along a straight track to airfeld B because at all times throughout the fight, the aeroplane is within glide range of a suitable forced landing location. Consequently, small piston engine aeroplanes should be fown at their maximum altitudes so that direct routes can be achieved. However, when planning the altitude, another regulation must be borne in mind. JAR-OPS 1 states, that when complying with the safe forced landing rule the aeroplane must not be assumed to be fying, with the engine operating at maximum continuous power at an altitude exceeding that at which the aeroplanes rate of climb equals 300 feet per minute. What this rules efectively does, is to limit the maximum altitude that can be used in order to comply with the forced landing rule. An aeroplane may operate at higher altitude than this regulation prescribes but the operator may not use the higher altitude in his calculation of glide range to a safe landing area. There is one last detail to be considered about complying with the forced landing rule. JAR-OPS 1 states, that in order to comply with the safe forced landing rule the assumed en-route gradient shall be the gross gradient of descent, increased by a gradient of 0.5%. In our example we had a gradient of descent of 7%, unfortunately though, as we have seen, the regulations do not permit this gradient to be used in our calculation. The gradient must be increased by 0.5% to 7.5% as shown in fgure 8.4304Chapter 8Single Engine Class BEn Route and DescentFigure 8.4 An illustration showing an example descent range calculationThis will lower the assumed descent performance of the aeroplane. The new deteriorated gradient that the regulations insist that we use is called the net gradient, and in our case this is 7.5%. If this net gradient has to be used the glide distance will reduce to 22 nm. So, although the aeroplane may achieve 23.5 nm following an engine failure, it must be assumed to glide only 22 nautical miles. Looking at fgure 8.5, this means that the circles around each safe landing area must reduce to a radius of 22 nm. Figure 8.5 An illustration showing the net track that is required to remain withnet glide range of each safe forced landing area305Chapter 8Single Engine Class BEn Route and DescentThus the planned track will have to change slightly so that at all points along the route the aeroplane is no further than 22 nm from a safe forced landing area. The track shown in fgure 8.5 meets the entire set of requirements as stated in JAR-OPS 1.INFORMATION IN CAP 698It was mentioned at the beginning of the chapter that CAP 698 has the en-route regulations in it for you, but that they were scatered through the manual and not conveniently collected in one place.On page 1 of section 2 in the general requirement paragraph, point c) is the regulation about ensuring that the aeroplane is not operated unless surfaces are available which permit a safe force landing to be carried out in the event of engine failure. However, the other regulations about complying with this rule are a few pages further on you will fnd the remaining en-route requirements. At the botom of page 8 of section 2. The frst part of the regulations to be found here, states that the aeroplane may not be assumed to be fying above the altitude at which a rate of climb of 300 feet per minute can be achieved. Underneath that rule you can the see the requirement which states that the net gradient of descent, in the event of engine failure, is the gross gradient + 0.5%.Although the concepts of range and endurance have been covered in a previous lesson, it is important for the pilot to be able to use the information in the aeroplane fight manual so that he may calculate the range and endurance of the aeroplane. In the aircraft manual there are graphs and tables which help you calculate the endurance and range of your aeroplane under various conditions. 306Chapter 8Single Engine Class BEn Route and DescentTYPICAL RANGE AND ENDURANCE GRAPHSShown in fgure 8.6 is a typical endurance graph for a light single engine class B aeroplane. As an example let us assume a cruise altitude of 7000ft with an outside air temperature of 7C at 65% power. Working through the graph hopefully you see that the endurance of the aircraft allowing 45 minutes of reserve fuel is 6.6 hours, or 7.4 hours allowing for no reserves.Figure 8.6 A typical endurance graph for a light single engine aeroplane307Chapter 8Single Engine Class BEn Route and DescentFigure 8.7 is for calculating the aircraft range and it works in exactly the same way as the endurance graph except that instead of working out airborne time, distance fown is shown.Figure 8.7 A typical endurance graph for a light single engine aeroplaneTo achieve these range and endurance fgures, be careful to follow the techniques described in the manual, especially with regard to correctly leaning the mixture.308Chapter 8Single Engine Class BEn Route and DescentQUESTIONS1. Which of the following statements correctly describes one of the general requirements about the operation of Class B single engine aeroplanes in the public transport category?a. They may fy at night.b. They must be fown so that an airfeld can be reached following engine failure.c. They are not to operate in IMC, except under special VFR.d. They must not be operated over water.2. According to the information in a light aircraft manual, which gives two power-setings for cruise, 65% and 75%. If you fy at 75% instead of 65%:a. Cruise speed will be higher and SFC will be the same.b. Cruise speed will be higher and SFC will be lower.c. Cruise speed will be higher and SFC will be higher.d. Cruise speed will be the same and SFC will be the same.3. According to the information in a light aircraft manual, which gives two power setings for cruise, 65% and 75%. If you fy at 65% instead of 75%:a. Endurance will be higher and SFC will be the same.b. Endurance will be higher and SFC will be lower.c. Endurance will be higher and SFC will be higher.d. Endurance will be the same and SFC will be the same.4. For the purpose of ensuring compliance with the en-route regulations, up to what maximum altitude is the aeroplane assumed to operate?a. The altitude where the rate of climb falls to 300 ft/min with maximum continuous power set.b. With maximum take-of power set, the altitude where the rate of climb exceeds 300 ft/min.c. With maximum continuous power set, the altitude where the rate of climb exceeds 300 ft/min.d. The altitude where the rate of climb increases to 300 ft/min with maximum take-ofpower set.

5. For the purpose of ensuring compliance with the en-route regulations, the en-route descent gradient must be:a. the gross gradient of descent decreased by 0.5%.b. the net gradient of descent decreased by 0.5%.c. the gross gradient of descent increased by 0.5%.d. 0.5%.6. To ensure a piston engine class B aeroplane can glide the furthest distance following engine failure, what speed must be fown?a. VMPb. 1.32 VMDc. 0.76 VMDd. VMD309Chapter 8Single Engine Class BEn Route and Descent7. Following engine failure, to maximise the descent range of a small piston engine aeroplane the aeroplane must be fown at:a. the speed for the maximum lift over drag ratiob. VMPc. the speed for minimum lift over drag ratio.d. a speed equal to the aeroplanes best angle of climb.310Chapter 8Single Engine Class BEn Route and DescentANSWERS1. C2. C3. B4. A5. C6. D7. A311Chapter 9Single Engine Class BLandingCHAPTER NINESINGLE ENGINE CLASS B - LANDINGContentsLANDING REQUIREMENT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313EXAMPLE LANDING REQUIREMENT . . . . . . . . . . . . . . . . . . . . . . . . . 313FACTORS TO BE ACCOUNTED FOR. . . . . . . . . . . . . . . . . . . . . . . . . . . 314CORRECTION FACTORS. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315DESPTACH RULES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315REFERENCE LANDING SPEED (VREF) . . . . . . . . . . . . . . . . . . . . . . . . . . 317PRESENTATION OF LANDING DATA / USING THE GRAPHS . . . . . . . . . . . 317QUESTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318ANSWERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322312Chapter 9Single Engine Class BLanding313Chapter 9Single Engine Class BLandingLANDING REQUIREMENTThis lesson will focus on the Class B single engine performance requirements for the landing stage of fight. There is actually only one regulation requirement for landing a single engined Class B aeroplane. This requirement is that the landing distance must not exceed the landing distance available. In other words, the aeroplane must be able to land within the length of the runway. In order to comply with this regulation, certain points must be taken into account. These will all be discussed in this lesson.As with other regulations we have dealt with in this course, the requirements relevant to the landing of single engined class B aeroplanes can be found in CAP 698. JAR-OPS 1 states that an operator must ensure that the landing mass of the aeroplane, for the estimated time of arrival, allows a full stop landing from 50 ft above the threshold within 70% of the landing distance available at the destination aerodrome and at any alternate aerodrome. This means that the aeroplane must be able to land within 70% of the landing distance available. The factor to use for such calculation is 1.43.EXAMPLE LANDING REQUIREMENTTo help clarify this, let us work through an example and use fgure 9.1. If the landing distance available at the destination airfeld is 2,200 ft long, then obviously the aeroplane must be able to land within 70% of 2,200 ft. To calculate this value, we must divide 2,200 ft by a factor of 1.43. Carrying out this simple calculation, gives us an answer of 1,538 ft. 1,538 ft is 70% of 2,200 ft. Therefore, the aeroplane must be able to achieve a full stop landing within 1,538 ft. Figure 9.1 An illustration showing that the aeroplane must demonstrate it can cometo a full stop within 70% of the landing distance availableThere is another way to look at the landing requirement. If the landing distance of the aeroplane is calculated to be 1,200ft, what is the minimum length of the landing distance available which will allow a pilot to comply with the 70% rule? Use fgure 9.2 to help. In this case we simply multiply 1,200 ft by 1.43. This gives us an answer of 1,716 ft. Therefore, the landing distance 314Chapter 9Single Engine Class BLandingavailable must be at least 1,716 ft long. If the destination aerodrome has a runway with a landing distance available in excess of 1,716 ft, the aeroplane will be able to land within 70% of the landing distance available and therefore satisfy the regulation requirement.Figure 9.2 The obtain the minimum length of runway required,multiply the gross landing distance by 1.43The 70% landing regulation helps us to see the diference between net and gross performance. Here the gross performance is 1,200 ft, but the net performance, which is always worse than the gross performance and is the one in a million worse case scenario, is 1,716ft. The landing regulation we have just been discussing can be found in CAP 698 in section 2 at the top of page 9. The regulation states that the landing distance, from a screen height of 50 ft, must not exceed 70% of the landing distance available. If the aeroplane cannot come to a full stop within this length, then the landing distance must be reduced, either by selecting a higher fap seting or simply by reducing the mass of the aeroplane.FACTORS TO BE ACCOUNTED FORThe regulations in CS-23 state that when calculating the gross landing distance, certain details must be accounted for. These are listed below. The gross landing distance shall take account of: the pressure altitude at the aerodrome standard temperature the runway surface conditions and the type of runway surface the runway slope not more than 50% of the reported headwind component or not less than 150% of the reported tailwind component the despatch rules for scheduled or planned landing calculations JAROPS 1.550 paragraph c.315Chapter 9Single Engine Class BLandingCORRECTION FACTORSHaving discussed the regulation requirements detailing the maximum landing distance that must be available at a destination airfeld or alternate airfeld, we will now examine in detail the regulations governing how the gross landing distance is calculated given non standard conditions like grass runways, wet runways and sloping runways.GRASSIn CAP 698 section 2, at the top of page 9 under point (b) we read that if the runway is of grass up to 20 cm high, the landing distance should be multiplied by a factor of 1.15. This would efectively increase the landing distance by 15%. WETIf there is an indication that the runway may be wet at the estimated time of arrival, the landing distance should be multiplied by a factor of 1.15. Again, multiplying the calculated landing distance by a factor of 1.15 will increase the landing distance by 15%. If the Aeroplane Manual gives additional information on landing on wet runways, this may be used even if it gives a lesser distance than that from the above paragraph.SLOPEThe landing distance should be increased by 5% for each 1% downslope. This means that for a 1% downslope the landing distance should be multiplied by a factor of 1.05. Therefore for a 2% downslope the factor would be 1.1. Point (d) also states that no allowance is permited for upslope. The reason for this is that upslope will reduce the landing distance. If a pilot were to ignore the reduction in the landing distance then a margin of safety would be incorporated into the landing distance calculation.DESPTACH RULESLastly, point (e) states that there must be compliance with the despatch rules for scheduled or planned landing calculations and that these can be found in CSOPS 1.550 paragraph (c).The despatch rules, found in CS-OPS 1.550 paragraph (c), state that for despatching an aeroplane, it must be assumed that: The aeroplane will land on the most favourable runway at the destination airfeld in still air, and, The aeroplane will land on the runway most likely to be assigned considering the probable wind speed and direction.If this second assumption cannot be met, the aeroplane may be despatched only if an alternate aerodrome is designated which full compliance of the regulatory despatch requirements can be met.These despatch regulations may sound a litle complicated, so lets examine them further.The frst assumption is that the aeroplane will land on the most favourable runway, in still air or zero wind. This means that, assuming zero wind at the destination airfeld, the runway which would accommodate the largest possible landing mass would be selected since that is the most favourable runway. 316Chapter 9Single Engine Class BLandingUsing an example which you see in fgure 9.3, frstly we try and understand the frst rule. Assuming zero wind and focusing on the column STILL AIR MASS, runway 04 and runway 22 both allow a maximum landing mass of 1,500 kg but runway 31 and runway 13, being much shorter runways, allow a maximum landing mass of only 1,000 kg. Therefore, the most favourable runway in still air is either runway 04 or 22, which both allow a maximum landing mass of 1,500 kg. These two masses have been highlight in red.Figure 9.3 An illustration showing example landing masses of an aeroplane basedupon no wind and the forecast wind conditionThe second despatch assumption is that the aeroplane will actually land on the most likely runway to be assigned considering the probable wind speed and direction. Now focus on the second column FORECAST WIND MASS. In this case, runway 04 has a strong headwind for the landing which will allow the aeroplane to land at a much greater mass than if there were no wind. In our example the headwind has increased the maximum landing mass to 1,750 kg. However, runway 22 has a tailwind, which will decrease the maximum landing mass to 800 kg. Runway 31 and runway 13 both have a full cross wind; therefore, the maximum landing mass for these two runways will be the same as for still air. If the second despatch assumption is that the aeroplane will land on the runway most likely to be assigned considering the probable wind speed and direction, then it would have to be runway 04 with a maximum landing mass of 1750 kg. This mass is highlighted in red.In summary then, the frst assumption requires that the pilot select runway 04 or 22 for a maximum still air landing mass of 1,500 kg, and the second assumption will require the pilot to select runway 04 for a maximum landing mass of 1,750 kg. Now we need to consider what to use as the despatch mass of the aeroplane. If the aeroplane was despatched at 1,750 kg mass, and then on arrival at the airfeld, the wind was less than 15 knots, the aeroplane would not be able to land. Therefore, the aeroplane must be despatched with a mass of 1,500 kg. Despatching the aeroplane with this mass would mean that, no mater what wind conditions prevail at the destination airfeld, the aeroplane will have a mass which will allow it to land at the airfeld.317Chapter 9Single Engine Class BLandingThe way to simplify the despatch rule is always consider the greatest mass in still air and the greatest mass in the forecast wind conditions and, of the two, take the lesser mass as the despatch mass. Of the two, it is the still air mass that is usually the lesser mass, as shown in our example. The exception is when there is a tailwind on a unidirectional runway. A unidirectional runway is a runway whose direction for take-of and landing is fxed in one direction. And in this case the maximum landing mass in the tailwind will be less than the maximum landing mass in still air.REFERENCE LANDING SPEED (VREF)You may recall that the regulatory speed at the landing screen height is called VREF, and, for a Class B single engine aeroplane it had to be no less than 1.3 times the stall speed in the landing confguration, (1.3 VSO). A pilot must adhere to the VREF speeds because they are the speeds which have been used to construct the landing graphs or table in the aeroplane fight manual. If a pilot were to deviate from these speeds, the required aircraft performance would not be achieved.PRESENTATION OF LANDING DATA / USING THE GRAPHSThis part of the chapter will deal with how the aeroplanes landing distance can actually be calculated. All aeroplanes will have either a pilot operating handbook or an aeroplane fight manual. The purpose of these manuals is to show not only how to operate the aeroplane but also to detail the aeroplanes performance. The example graph we will use is fgure 2.4 on page 10 of section 2 in CAP 698. Always take a look at the associated conditions frst, paying particular atention to the power and fap setings as shown here. Also notice that this graph assumes a runway which is paved, level and dry. If the runway conditions required for a given calculation are diferent, corrections will need to be made to the values that this graph will give. We saw these correction factors earlier. The left hand carpet of the graph involves the variations in temperature and pressure altitude. This part of the graph accounts for the efect of air density on the landing distance. The middle carpet account for the efect of the mass and to the right of this carpet is the wind correction carpet. Notice the diferences in the slope of the headwind and tailwind lines. This means the 150% and 50% wind rules have been applied. The last carpet on the far right of the graph is labelled obstacle height. Although there is no obstacle as such you will recall that the landing starts at a height of 50 ft above the landing surface.Follow through the example that has been carried out for in the graph. This will help you to use the graph correctly. Not only is there an example on the graph itself, but if you look at the botom of page 9 of section 2 in CAP 698 you will see another example which you can work through. Use the examples; they are there to help you. If you need practice on working through the graphs use the questions at the end of the chapter.318Chapter 9Single Engine Class BLandingQUESTIONS1. For this question use Performance Manual CAP 698 SEP 1 Figure 2.4With regard to the landing chart for the single engine aeroplane determine the landing distance from a height of 50 ft. Given:-O.A.T: 27 CPressure Altitude: 3,000 ftAeroplane Mass: 2,900 lbsTailwind component: 5 ktFlaps: Landing position (down)Runway: Tarred and Drya. approximately : 1,120 feetb. approximately : 1,700 feetc. approximately : 1,370 feetd. approximately : 1,850 feet2. For this question use Performance Manual CAP 698 SEP 1 Figure 2.4With regard to the landing chart for the single engine aeroplane determine the landing distance from a height of 50 ft.

a. 0.70b. 1.67c. 1.43d. 0.6011. The landing feld length required for single engine class B aeroplanes at the alternate aerodrome is the demonstrated landing distance plusa. 92%b. 43%c. 70%d. 67%321Chapter 9Single Engine Class BLanding12. The actual dry landing distance of a single engine class B aeroplane is calculated at 1,300 ft. What is the minimum landing distance to be required at the aerodrome in order to comply with the landing regulations?. (runway is wet at the estimated time of arrival) a. 1,495 ftb. 2,138 ftc. 1,859 ftd. 1,130 ft322Chapter 9Single Engine Class BLandingANSWERS1. D2. D3. B4. D5. C6. C7. A8. B9. B10. A11. B12. B323Chapter 10Multi Engine Class BTake OfCHAPTER TENMULTI-ENGINED CLASS B - TAKE OFFContentsTAKE-OFF REQUIREMENTS. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325GRADIENT REQUIREMENT JAR-OPS 1.530 . . . . . . . . . . . . . . . . . . . . . . 325FIELD LENGTH REQUIREMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325FACTORS TO BE ACCOUNTED FOR. . . . . . . . . . . . . . . . . . . . . . . . . . . 326SURFACE CONDITION FACTORS . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326PRESENTATION OF TAKE-OFF DATA. . . . . . . . . . . . . . . . . . . . . . . . . . 327ACCELERATE-STOP DISTANCE REQUIREMENTS . . . . . . . . . . . . . . . . . . 327TAKE-OFF SPEEDS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327OBSTACLE CLEARANCE REQUIREMENTS (JAR-OPS 1.535) . . . . . . . . . . . . 328TAKE-OFF FLIGHT PATH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 328CONSTRUCTION OF THE FLIGHT PATH. . . . . . . . . . . . . . . . . . . . . . . . 328QUESTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331ANSWERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334324Chapter 10Multi Engine Class BTake Of325Chapter 10Multi Engine Class BTake OfTAKE-OFF REQUIREMENTSThe take-of requirements for multi-engined Class B aircraft (other than those in the commuter category) are the same as for single engined aircraft except that multi engine class B aeroplanes must additionally demonstrate a minimum climb gradient performance and an obstacle clearance capability.GRADIENT REQUIREMENT JAR-OPS 1.530Three climb gradient requirements must be considered, and the most limiting will determine the maximum permissible mass. You can fnd these in CAP 698 under paragraph 3.1.2 on page 9 of section 3. ALL ENGINE OPERATINGA minimum climb gradient of 4% is required with: take-of power on each engine landing gear extended, except that if the landing gear can be retracted in not more than 7 seconds, it may be assumed to be retracted the wing faps in the take-of position a climb speed of not less than the greater of 1.1VMC and 1.2VSI ONE ENGINE INOPERATIVEThe climb gradient at an altitude of 400ft above the take-of surface must be measurably positive with: the critical engine inoperative and its propeller in the minimum drag position. the remaining engine at take-of power the landing gear retracted the wing faps in the take-of position a climb speed equal to that achieved at 50 ft andThe climb gradient must not be less than 0.75% at an altitude of 1500 ft above the take-of surface, with: the critical engine inoperative and its propeller in the minimum drag position the remaining engine at not more than maximum continuous power the landing gear retracted the wing faps retracted a climb speed not less than 1.2 VSIFIELD LENGTH REQUIREMENTSThe Field Length Requirements are the same as for the single engine aeroplane and are detailed below. You can also fnd them in CAP 698 under paragraph 2.1.1 of page 1 and 2 of section 3. When no stopway or clearway is available, the take-of distance when multiplied by 1.25 must not exceed TORA (Gross TOD x 1.25 must not exceed the TORA)326Chapter 10Multi Engine Class BTake Of When a stopway and/or clearway is available the take-of distance must: not exceed TORA (Gross TOD must not exceed the TORA) when multiplied by 1.3, not exceed ASDA (Gross TOD x 1.3 must not exceed the ASDA) when multiplied by 1.15, not exceed TODA (Gross TOD x 1.15 must not exceed the TODA)FACTORS TO BE ACCOUNTED FORThe regulations in CS-23 state that when calculating the gross take-of distance, in other words, before we add the factors mention previously, certain details must be accounted for. These are listed below. The gross take-of distance required shall take account of: the mass of the aeroplane at the start of the take-of run the pressure altitude at the aerodrome the ambient temperature at the aerodrome the runway surface conditions and the type of runway surface the runway slope not more than 50% of the reported headwind component or not less than 150% of the reported tailwind component SURFACE CONDITION FACTORSFrom the list above the regulations stipulate that due account must be made of the runway condition. Most performance data in the aeroplane fight manual assumes a level, dry and hard runway. Therefore, correction factors must be applied to the gross take-of distance when the runway conditions are diferent. There are various correction factors such as for grass runways, wet runways and runways which are sloped. These factors are detailed next.At the top of page 2 section 2 under paragraph c) in CAP 698 it states that if the runway is other than dry and paved the following correction factors must be used when determining the take-of distance. These are shown for you in fgure 6.2Figure 10.1 The corrections factors that need to be applied to the gross take-of distance when the runway is other than dry, paved and level.As you have learnt already, grass runways will increase the take-of distance compared to paved runways. Here the factor to use to account for the efect of dry grass is 1.2 and 1.3 if the grass is wet.327Chapter 10Multi Engine Class BTake OfAt the top of page 2 section 2 underneath the table, point d) details the corrections to be applied if there is a slope to the runway. It states that a pilot must increase the take-of distance by 5%, or by a factor of 1.05, for ever 1% upslope. However, it also states that no factorisation is permited for downslope. In other words, when an aeroplane may be taking-of on a downwards sloping runway no correction factor is to be applied for the downslope. The reason for ignoring the downslope is because a downslope will decrease the take-of distance. This helps to add a litle extra safety to the take-of distance calculation.PRESENTATION OF TAKE-OFF DATAShown in CAP 698 on page 3 and 7 of section 3 you can see the take-of distance graphs for a typical Class B multi engine aeroplane. The one page is normal take-of and the one on page 7 is a maximum efort take-of, in other words a short feld take-of. There are several examples in CAP 698, one at the botom of page 2 and the other at the top of page 6 in section 3. Ensure you go through these examples to help you with your graph work and to help to when to apply the various factors.ACCELERATE-STOP DISTANCE REQUIREMENTSOther than for commuter category aircraft, there is no requirement for accelerate-stop distance, but the data may be given. Shown in CAP 698 on page 5 and 8 of section 3 you can see the accelerate stop distance graphs for a typical Class B multi engine aeroplane. Similarly with other graphs, there are examples which you should go through.For commuter category aircraft, the accelerate-stop distance is the sum of the distances necessary to : Accelerate the aircraft to VEF with all engines operating. Accelerate from VEF to V1 assuming the critical engine fails at VEF Come to a full stop from the point at which V1 is reached.TAKE-OFF SPEEDSOther than for commuter category aircraft, these are the same as for the single engined ClassB aircraft. The gross take-of distance required is the distance from the start of take-of to a point 50 ft. above the take-of surface, with take-of power on each engine, rotating at VR and achieving the specifed speed at the screen.VR The rotation speed, must not be less than : 1.05 VMC VS1The speed at 50 ft. must not be less than (the take-of safety speed): a speed that is safe under all reasonably expected conditions VMC VS1VMC for take-of must not exceed 1.2 VS1These speeds are not found in CAP 698 and must be commited to memory.328Chapter 10Multi Engine Class BTake OfOBSTACLE CLEARANCE REQUIREMENTS (JAR-OPS 1.535)Multi-engined Class B aircraft must demonstrate clearance of obstacles after take-of up to a height of 1500ft. All the obstacle clearance requirements can be found in CAP 698 and therefore they do not need to be learnt. You can fnd them all on page 9 of section 3.Obstacles must be cleared by: a vertical margin of at least 50ft or a horizontal distance of at least 90m + 0.125 D where D is the distance from the end of the TODA, or the end of the TOD if a turn is scheduled before the end of the TODA. For aeroplanes with a wingspan of less that 60m the horizontal distance may be taken as 60m + half the wingspan + 0.125D.The following conditions must be assumed: the fight path begins at a height of 50 ft above the surface at the end of the TOD required and ends at a height of 1500 ft above the surface. the aeroplane is not banked before it has reached the height of 50 ft, and thereafter that the angle of bank does not exceed 15. failure of the critical engine occurs at the point on the all engine take-of fight path where visual reference for the purpose of avoiding obstacles is expected to be lost. the gradient to be assumed from 50 ft to the point of engine failure is equal to the average all engine gradient during climb and transition to the en-route confguration, multiplied by a factor of 0.77 the gradient from the point of engine failure to 1500 ft. is equal to the one engine inoperative en-route gradient.If the fight path does not require track changes of more then 15, obstacles do not need to be considered if the lateral distance is greater than 300m if in VMC or 600m for all other conditions. If the fight path requires track changes of more than 15, obstacles need not be considered if the lateral distance is greater than 600m in VMC or 900m for all other conditions.TAKE-OFF FLIGHT PATHThe fight path profle performance should take account of: the mass of the aeroplane at the commencement of the take-of run the pressure altitude at the aerodrome the ambient temperature not more than 50% of the reported headwind component and not less than 150% of the reported tailwind component.CONSTRUCTION OF THE FLIGHT PATHThe fight path profle will depend on whether or not visual reference is lost before reaching1500 ft. VISIBILITY CLEAR TO 1500 ft Determine the TOD required for the take-of mass Determine the all engines net gradient (gross gradient x 0.77)329Chapter 10Multi Engine Class BTake Of Divide the height gain (1450 ft) by the gradient to determine the distance travelled (feet) from 50 ft to 1500 ft. The profle may be ploted as shown in fgure 10.3 and clearance of obstacles assessed.Figure 10.3Figure 10.2. The obstacle clearance climb profle if there is no cloudAlternatively for a single obstacle, fnd the TOD req. and gradient as above, then multiply the distance from the TODR to the obstacle by the gradient to fnd the height gain, and add 50 ft to fnd the aeroplane height at the obstacle distance. This must exceed the obstacle height by 50 ft.If the obstacle is not cleared by 50 ft, a lower take-of mass must be assumed and a revised height calculated. The maximum mass which will just clear the obstacle by 50 ft can then determined by interpolation. CLOUD BASE BELOW 1500 ftIf visual reference is lost before 1500 ft, the fight path will consist of two segments.Segment 1 (From 50 ft to cloud base)Distance from 50 ft to cloud base = height gain all engine net gradient x 100Height gain = cloud base - 50 ft Segment 2 (From cloud base to 1500 ft)Distance = height gain Gross gradient with one engine inoperative x 100The profle may be ploted as shown in Figure 10.3 and clearance of obstacles assessed. If the required clearance is not achieved a reduced take-of mass must be assumed and a second fight path calculated. As before the maximum permissible weight may be determined by interpolation.330Chapter 10Multi Engine Class BTake OfFigure 10.4Figure 10.3 The obstacle clearance climb profle if there is cloud.If the climb data is given in terms of rate of climb, this can be converted to gradient:Gradient% = Rate of climb(ft/min) Aircraft true groundspeed x 100Alternatively the time on each segment can be calculated:Time (mins) = height gain (ft) Rate of climb (ft/min)and the distance on each segment obtained from:Distance (ft) = Aircraft True Ground Speed (ft/min) X Time (min)331Chapter 10Multi Engine Class BTake OfQUESTIONS 1. For a Class B multi-engine aeroplane at an aerodrome with no stopway or clearway, the length of take-of run that must be available for take-of, to satisfy the requirements:a. Must not be less than the gross take of distance to 50ftb. Must not be less than 1.15 times the gross take of distance to 50ftc. Must not be less than 1.25 times the gross take of distance to 50ftd. Must not be less than 1.3 times the gross take of distance to 50ft2. For a multi-engine Class B aircraft, the rotation speed VR:a. Must not be less than either 1.1 VS1 or 1.05 VMCb. Must not be less than VS1c. Must not be less than 1.05 VS1 or 1.1 VMCd. Must not be less than VMC3. For a multi-engine Class B aircraft, the take-of safety speed must:a. not be less than either 1.1 VS1 or 1.05 VMCb. greater than 1.2 VMC or 1.1 VS1c. not be less than either 1.2 VS1 or 1.1 VMCd. greater than VS14. For this question use Performance Manual CAP 698 MEP 1 Figure 3.4.Determine the accelerate-stop distance from brake release to a full stop given an abort speed of 64 KIAS and a reaction time of three seconds.Given:- O.A.T : 27C Pressure Altitude: MSL Aeroplane Mass: 3,750 lbs Tailwind component: 5 kt Flaps: 25 Runway: Paved, Level and Drya. 2,200 ftb. 1,800 ftc. 3,300 ftd. 2,400 ft332Chapter 10Multi Engine Class BTake Of5. For this question use Performance Manual CAP 698 MEP 1 Figure 3.2.Determine the maximum permissible mass that will allow the aeroplane to come to full stop given an accelerate-stop distance available of 3,200 ft.Given:- O.A.T : ISA Pressure Altitude: MSL Headwind component: 5 kt Flaps: 0 Runway: Paved, Level and Drya. 3,550 lbsb. 4,100 lbsc. 4,250 lbsd. 3,000 lbs6. For this question use Figure 3.2 in CAP 698. With regard to the graph for the light twin aeroplane, will the accelerate and stop distance be achieved in a take-of where the brakes are released before Take-of power is set?a. It does not mater which take-of technique is being used.b. No, the performance will be worse than in the chart.c. Performance will be beter than in the chart.d. Yes, the chart has been made for this situation.7. When assessing obstacle clearance after take-of for a twin engined Class B aircraft, the climb from 50 feet to 1,500 feet:a. Is always assumed to take place with all engines operating.b. Assumes that an engine fails at the point where visual reference of the obstacle is lost.c. Always assumes that an engine has failed at 50 feet.d. Assumes that an engine fails at 400 feet above ground level.8. A light twin engine aircraft is climbing from the screen height of 50ft, and has an obstacle 10,000m along the net fight path. If the net climb gradient is 10%, there is no wind and the obstacle is 900m above the aerodrome elevation then what will the clearance be?a. The aircraft will not clear the object.b. 85 mc. 100 md. 115 m9. By what vertical margin must a multi-engine class B aeroplane clear an obstacle in the take-of fight path?a. 35 ft.b. 50 ftc. There is no obstacle clearance requirementd. 60 m + 0.125D333Chapter 10Multi Engine Class BTake Of10. Regarding the take-of climb requirements for a multi-engine class B aeroplane, what is the minimum all engine climb gradient after take-of?a. 0.75%b. >0%c. 4%d. 2.4%11. If a multi-engine class B aeroplane is unable to achieve the required vertical clearance over an obstacle, by what minimum horizontal margin must the obstacle by cleared (assume wing span < 60 m) a. 60 m + 1/2 wingspan + 0.125Db. 90 m + 0.125Dc. 60 m / D + 0.125d. There is no minimum horizontal clearance requirement.12. What is the maximum bank angle permited within the take-of fight path up to 1,500 ft for a multi-engine class B aeroplane?a. 25b. 10c. 5d. 1513. By what regulatory factor must the all engine climb gradient of a class B multi engine aeroplane be multiplied by in order to comply with the obstacle clearance requirements?a. 0.5%b. 0.5c. 0.77d. 1.43334Chapter 10Multi Engine Class BTake OfANSWERS1 C2 A3 C4 D5 B6 B7 B8 D9 B10 C11 A12 D335Chapter 11Multi Engine Class BEn Route and DescentCHAPTER ELEVENMULTI ENGINED CLASS B - EN-ROUTE AND DESCENTContentsEN-ROUTE REQUIREMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337THE DRIFT DOWN . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338CONSTRUCTION OF THE DRIFT DOWN PROFILE . . . . . . . . . . . . . . . . . . 339QUESTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341ANSWERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342336Chapter 11Multi Engine Class BEn Route and Descent337Chapter 11Multi Engine Class BEn Route and DescentEN-ROUTE REQUIREMENTSThis chapter will focus on the Class B multi engine performance requirements for the en-route and descent stage of fight. The en-route part of the fight is considered to be from 1500 ft above the airfeld from which the aeroplane has taken of, to 1000 ft above the destination airfeld.As with all the other requirements we have come across, the en-route and descent requirements are writen in JAR-OPS 1 and abbreviated into CAP 698. However, the abbreviated version in CAP 698 is very clear and almost word for word like the original requirements published in JAR-OPS 1. Looking at CAP 698 at the top of page 17 of section 3 you can see the en-route requirements. The frst sentence simply reminds you what the en-route stage of fight is. These requirements are listed below. An operator shall ensure that the aeroplane, in the meteorological conditions expected for the fight, and in the event of the failure of one engine, with the remaining engines operating within the maximum continuous power conditions specifed, is capable of continuing fight at or above the relevant minimum altitudes for safe fight stated in the Operations Manual to a point 1000 ft above an aerodrome at which the performance requirements for landing can be met.This requirement is almost identical to the single engine aeroplane. The only diference being that whereas the single engine aeroplane has to be capable of landing in a suitable feld after engine failure, the multi engine aeroplane must be capable of a higher performance level and therefore continue fight and land at a suitable airfeld. Therefore, in the event of engine failure, the multi engine aeroplane should have a level of performance such that it can, even with an engine failure, get to an airfeld to land. However, as with other requirements we have covered, the en-route requirements do have some compliance rules. In fact, these compliance rules are very similar to the ones we mentioned in the single engine en-route lesson.In order for the pilot to be able to abide by the regulation, the descent range with one engine in-operative must be known as well as the one engine in-operative cruise range. Once these have been calculated, a fight track must be ploted that will ensure an airfeld is always within the total one engine inoperative descent distance. Herein lies the compliance rules because the compliance rules relate to how that descent range is calculated. When showing compliance with the rules shown above. The aeroplane must not be assumed to be fying at an altitude exceeding that at which the rate of climb equals 300 ft/min with all engines operating within the maximum continuous power conditions specifed; and The assumed en-route gradient with one engine inoperative shall be the gross gradient of descent or climb, as appropriate, respectively increased by a gradient of 0.5%, or decreased by a gradient of 0.5%.What the frst compliance rule means is the aeroplane must not use the extra altitude above the 300 feet per minute altitude to gain extra range to help comply with landing at an airfeld after engine failure.The compliance rule is saying that when calculating the descent range to work out if the aeroplane can make it to an airfeld, the gross gradient of descent must be increased by 0.5%. This adjusted gradient is the net gradient and it is simply adding a safety margin into the aeroplanes descent range. In this case an airfeld must be within the net descent range and not the gross descent range.338Chapter 11Multi Engine Class BEn Route and DescentTHE DRIFT DOWNCalculating the descent range of a twin engine aeroplane after engine failure is not as easy as it was for the single engine aeroplane. For the single engine aeroplane, to calculate the descent range it was simply the height of the aeroplane divided by the descent gradient and multiplied by 100. However, the gradient of descent of a twin engine aeroplane following engine failure is constantly changing. Let us explain why. In straight and level fight the forward force of thrust balances the rearward force of drag. When the engine fails, there is more rearward force than forward force, and, as a result of the excess drag the aeroplane will slow down if level fight is maintained. To maintain the speed, which should be kept at VMD, the thrust force generated by the remaining live engine must be augmented so that the forces can once again be balanced. Figure 11.1 The forces of an aeroplane in the early part of the driftdownThe only way to do this is to lower the nose so that weight can act forward and provide enough weight apparent thrust to balance the excess drag. If the nose is lowered by a sufcient amount then the forces will once again balance and VMD can be maintained. The only side efect is that the aeroplane is descending. However, as the aeroplane descends in the atmosphere, the air density increases. This means that the thrust being produced by the remaining engine increases which reduces the excess drag. Now that there is no need for so much weight apparent thrust since the excess drag has reduced. To reduce the amount of weight apparent thrust, the nose is raised a litle. This process can continue until the remaining engine generates sufcient thrust to balance the drag without any need for weight apparent thrust. At the altitude where this balance occurs the aeroplane is able to level of. In summary then, after engine failure in the cruise, the aeroplane is forced to descend, but as it descends the aeroplane can slowly reduced the descent angle until the aeroplane can once more fy level. This procedure is known as the Driftdown procedure and it produces a Driftdown fight profle similar to the one shown in fgure 11.2.339Chapter 11Multi Engine Class BEn Route and DescentFigure 11.2 The driftdown profle split into manageable segments foreasy calculation of the descent rangeCONSTRUCTION OF THE DRIFT DOWN PROFILEIt was stated that calculating the descent range for a twin engine aeroplane after engine failure was complicated, and the reason, which is now hopefully apparent, is that the descent gradient or descent angle is constantly changing. In the absence of a Driftdown graph then the only feasible way of calculating the descent range is to break down the profle into manageable segments and carry out several calculations, as shown in fgure 11.2.Each of these calculations will need the net descent gradient at that atitude and the vertical interval of that segment. This will give the horizontal distance covered for that segment. To fnd the descent range, simply add all the horizontal distances in all the segments. After the descent range has been calculated and the aeroplane is able to fy straight and level, the last remaining information that is needed is to fnd out the one engine inoperative cruise range. Once this is known, it can be added to the descent range of the Driftdown profle to give the total range of the aeroplane following engine failure. Therefore, at any point along the fight there must be an airfeld at which a landing can be made within the total range from engine failure. To ensure this, a circle, whose radius is the total single engine range, is drawn around each airfeld between the departure and destination points. To comply with the regulations, the aeroplane track must fall inside these circles. In doing so the aeroplane will comply with JAR-OPS 1 which states that in the event of engine failure the aeroplane is capable of continuing fight to an aerodrome where a landing can be made. 340Chapter 11Multi Engine Class BEn Route and Descent341Chapter 1Multi Engine Class BEn Route and DescentQUESTIONS1. For a multi-engine class B aeroplane, the en-route phase extends from:a. 1,000 ft above the take-of surface to 1,500 ft above the landing surface.b. 1,500 ft above the take-of surface to 1,000 ft above the landing aerodrome level.c. From the start of level fight to the end of level fight.d. 50 ft above the take-of surface to 1,500 ft above the landing aerodrome level.2. Following engine failure in the cruise, what is the name given to the descent procedure from the cruise altitude to the one engine inoperative ceiling?a. Descent profle.b. Descent procedure.c. Driftdown.d. Emergency descent.3. Why does the descent profle of the driftdown procedure steadily become shallower?a. As density increases, the remaining engine generates more thrust.b. Drag starts to decrease towards the end of the driftdown procedure.c. Weight apparent thrust decreases with increasing density.d. The increase in gravitational acceleration causes the weight apparent thrust to increase.4. Which of the statements below correctly describes on the en-route requirements for a multi-engine class B aeroplane?a. In the event of engine failure, the aeroplane is capable of reaching a place at which a safe forced landing can be made.b. In the event of the failure of one engine, the aeroplane is capable of continuing fight to an aerodrome.c. The aeroplane must not be operate unless a landing into safe forced landing areas can be made.d. The aeroplane cannot operate above an altitude where the rate of climb is less than 300 ft/min.5. Following engine failure in a class B multi-engine aeroplane at cruise altitude, why is the aeroplane forced to descend?a. There is insufcient oxygen at high altitude to support the passengers.b. Drag increases so that it exceeds the thrust available.c. The one engine inoperative ceiling is lower than the two engine operative ceiling.d. There is insufcient thrust to balance drag.6. For a multi-engine class B aeroplane, following engine failure, what speed should be used during the descent to the one engine inoperative ceiling?a. VMDb. VMPc. 1.32 VMDd. VX342Chapter 11Multi Engine Class BEn Route and DescentANSWERS1 B2 C3 A4 B5 D6 A343Chapter 12Multi Engine Class BLandingCHAPTER TWELVEMULTI-ENGINED CLASS B - LANDINGContentsLANDING REQUIREMENTS. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345LANDING CLIMB REQUIREMENTS / GRADIENT REQUIREMENT . . . . . . . . . 345LANDING DISTANCE REQUIREMENTS JAR-OPS 1.545 . . . . . . . . . . . . . . . . 346FACTORS TO BE ACCOUNTED FOR. . . . . . . . . . . . . . . . . . . . . . . . . . . . 347CORRECTION FACTORS. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347DESPTACH RULES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 348REFERENCE LANDING SPEED (VREF) . . . . . . . . . . . . . . . . . . . . . . . . . . . 348PRESENTATION OF LANDING DATA / USING THE GRAPHS . . . . . . . . . . . . 348QUESTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349ANSWERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352344Chapter 12Multi Engine Class BLanding345Chapter 12Multi Engine Class BLandingLANDING REQUIREMENTSThis chapter will focus on the Class B multi engine performance requirements for the landing stage of fight. There are two main regulation requirements.1) The frst requirement is that the landing distance must not exceed the landing distance available. In other words, the aeroplane must be able to land within the length of the runway. This requirement can be called the landing distance requirement.2) The second requirement is that should the aeroplane be unable to land, it must be able to climb away from the aerodrome with an adequate climb gradient. This later requirement can be called the landing climb requirement.As with other regulations we have dealt with, the requirements relevant to the landing of multi engined class B aeroplanes can be found in CAP 698. However, caution must be exercised since CAP 698 contains very abbreviated versions of the requirements and therefore may be misleading. CAP 698, is after all, only supposed to contain supplementary information for examination purposes only.LANDING CLIMB REQUIREMENTS / GRADIENT REQUIREMENTLets discuss the landing climb requirements frst. If, for whatever reason, a landing was not possible, then the aeroplane should have a level of performance that would enable it to climb safely away from the airfeld. This must be possible will either both engines operating or with one engine inoperative.The landing climb requirements originate from the certifcation specifcation rules in CS-23 but are adopted in a brief format into JAR-OPS 1. JAR-OPS 1 divides the landing climb requirements into, all engines operating, and one engine inoperative. You can fnd these requirements in CAP 698 at the botom of page 17 and the top of page 18 of section 3 so you dont need to commit these requirements to memory.ALL ENGINES OPERATING / BALKED LANDING REQUIREMENTWith all engines operating, the steady gradient of climb must be at least 2.5%. This gradient must be achieved with: The power developed 8 seconds after moving the power controls to the take-of position. The landing gear (undercarriage) extended. Flaps at the landing seting. Climb speed equal to VREF.

Note: In order to demonstrate this climb capability for certifcation and for operational purposes, the undercarriage is assumed to be extended and the wing faps in the landing position.346Chapter 12Multi Engine Class BLandingONE ENGINE INOPERATIVE / MISSED APPROACH REQUIREMENTSWith the critical engine inoperative, the gradient of climb must not be less than 0.75% at an altitude of 1500 feet above the landing surface. This gradient must be achieved with The critical engine inoperative and the propeller feathered The live engine set at maximum continuous power The landing gear (undercarriage) retracted The faps retracted Climb speed not less than 1.2 VS1. Notice this time that in order to demonstrate this climb capability for certifcation and for operational purposes, the undercarriage and wing faps are assumed to be retracted. The reason that the critical engine in-operative gradient requirement is much less than the all engine climb gradient requirement and that the aeroplane confguration is cleaner is because the failure of the critical engine results in an approximate 75% loss of climb gradient. Seting too high a level of regulation would impede on the operational capability of the aeroplane in terms of it payload because if the aeroplane is unable to atain these gradients, then the weight of the aeroplane must be reduced to an amount which can allow the gradient requirements to be met. The term used to describe the maximum mass that can be carried and still atain the minimum gradient is called the Landing Climb Limit Mass.The climb gradient requirements mentioned here are specifc to aeroplanes in the normal, utility and aerobatic category of more than 2,722 kg and therefore only represent a portion of the requirements for multi engine class B aeroplanes.An example of a landing climb performance graph is on page 19 of section in CAP 698. This graph is for the balked landing, in other words an all engine full power go around. However, notice that the graph only gives you a rate of climb. Therefore, in order to know if the aeroplane is achieving the minimum required gradient of 2.5% you must convert the rate of the climb into a gradient. An example of such a calculation is shown for you at the botom of page 18 of section 3 in CAP 698.LANDING DISTANCE REQUIREMENTS JAR-OPS 1.545The landing distance requirements for multi-engined Class B aircraft are the same as for single engined aircraft (see Chapter 9). You can see these requirements in CAP 698 in the middle of page 17 of section 3.JAR-OPS 1 states that an operator must ensure that the landing mass of the aeroplane, for the estimated time of arrival, allows a full stop landing from 50 ft above the threshold within 70% of the landing distance available at the destination aerodrome and at any alternate aerodrome.This means that the aeroplane must be able to land within 70% of the landing distance available. The factor to use for such calculation is 1.43.347Chapter 12Multi Engine Class BLandingFACTORS TO BE ACCOUNTED FORThe regulations in CS-23 state that when calculating the gross landing distance, certain details must be accounted for. These are listed below. The gross landing distance shall take account of: the pressure altitude at the aerodrome standard temperature the runway surface conditions and the type of runway surface the runway slope not more than 50% of the reported headwind component or not less than 150% of the reported tailwind component the despatch rules for scheduled or planned landing calculations JAROPS 1.550 paragraph c).CORRECTION FACTORSHaving discussed the regulation requirements detailing the maximum landing distance that must be available at a destination airfeld or alternate airfeld, we will now examine in detail the regulations governing how the gross landing distance is calculated given non standard conditions like grass runways, wet runways and sloping runways.GRASSIn CAP 698 section 3, in the middle of page 17 under point (b) we read that if the runway is of grass up to 20 cm high on frm soil, the landing distance should be multiplied by a factor of 1.15. This would efectively increase the landing distance by 15%. WETPoint c) states that if there is an indication that the runway may be wet at the estimated time of arrival, the landing distance should be multiplied by a factor of 1.15. Again, multiplying the calculated landing distance by a factor of 1.15 will increase the landing distance by 15%. If the Aeroplane Manual gives additional information on landing on wet runways, this may be used even if it gives a lesser distance than that from the above paragraph.SLOPEPoint d) states the landing distance should be increased by 5% for each 1% downslope. This means that for a 1% downslope the landing distance should be multiplied by a factor of 1.05. Therefore for a 2% downslope the factor would be 1.1. Point (d) also states that no allowance is permited for upslope. The reason for this is that upslope will reduce the landing distance. If a pilot were to ignore the reduction in the landing distance then a margin of safety would be incorporated into the landing distance calculation.348Chapter 12Multi Engine Class BLandingDESPTACH RULESLastly, point (e) states that there must be compliance with the despatch rules for scheduled or planned landing calculations and that these can be found in JAROPS 1.550 paragraph (c).The despatch rules, found in JAR-OPS 1.550 paragraph (c), state that for despatching an aeroplane, it must be assumed that;1) The aeroplane will land on the most favourable runway at the destination airfeld in still air, and, 2) The aeroplane will land on the runway most likely to be assigned considering the probable wind speed and direction.If this second assumption cannot be met, the aeroplane may be despatched only if an alternate aerodrome is designated which full compliance of the regulatory despatch requirements can be met.REFERENCE LANDING SPEED (VREF)You may recall that the regulatory speed at the landing screen height is called VREF, and, for a Class B single engine aeroplane it had to be no less than 1.3 times the stall speed in the landing confguration, (1.3 VSO). A pilot must adhere to the VREF speeds because they are the speeds which have been used to construct the landing graphs or table in the aeroplane fight manual. If a pilot were to deviate from these speeds, the required aircraft performance would not be achieved.PRESENTATION OF LANDING DATA / USING THE GRAPHSThis part of the chapter will deal with how the aeroplanes landing distance can actually be calculated. All aeroplanes will have either a pilot operating handbook or an aeroplane fight manual. The purpose of these manuals is to show not only how to operate the aeroplane but also to detail the aeroplanes performance. The example graph we will use is fgure 3.9 on page 21 of section 3 in CAP 698. Always take a look at the associated conditions frst, paying particular atention to the power and fap setings as shown here. Also notice that this graph assumes a runway which is paved, level and dry. If the runway conditions required for a given calculation are diferent, corrections will need to be made to the values that this graph will give. We saw these correction factors earlier. The left hand carpet of the graph involves the variations in temperature and pressure altitude. This part of the graph accounts for the efect of air density on the landing distance. The middle carpet account for the efect of the mass and to the right of this carpet is the wind correction carpet. Notice the diferences in the slope of the headwind and tailwind lines. This means the 150% and 50% wind rules have been applied. The last carpet on the far right of the graph is used because you will recall that the landing starts at a height of 50 ft above the landing surface. If you travel straight through the last carpet you will only calculate the landing roll, which in the example on the graph is 1,120 ft.Follow through the example that has been carried out for in the graph. This will help you to use the graph correctly. Not only is there an example on the graph itself, but if you look at the botom of page 20 of section 3 in CAP 698 you will see another example which you can work through. Use the examples; they are there to help you. If you need practice on working through the graphs use the questions at the end of the chapter.349Chapter 12Multi Engine Class BLandingQUESTIONS1. The Landing Distance Available at an aerodrome is 2,500 feet. For a Class B aircraft, what distance should be used in the landing distance graph to obtain the maximum permissible landing weight, if the runway has a paved wet surface with a 1% downhill slope?a. approximately : 1,665 ftb. approximately : 1,447 ftc. approximately : 1,748 ftd. approximately : 2,500 ft2. A runway is contaminated with 0.5 cm of wet snow. Nevertheless the fight manual of a light twin authorises a landing in these conditions. The landing distance will be, in relation to that for a dry runway:a. reduced.b. substantially decreased.c. increased.d. unchanged.3. For this question use Performance Manual CAP 698 MEP 1 Figure 3.9With regard to the normal landing chart for the multi engine aeroplane determine the landing distance from a height of 50 ft.

Given:-O.A.T : 30CPressure Altitude: 0 ftAeroplane Mass: 4,500 lbsHeadwind component: 10 ktFlaps: 40Runway: Paved, level and drya. approximately : 1,300 feetb. approximately : 2,050 feetc. approximately : 2,395 feetd. approximately : 2,475 feet4. For this question use Performance Manual CAP 698 MEP 1 Figure 3.9With regard to the graph for normal landing performance, what is the maximum allowable landing mass in order to comply with the landing regulations? Given:- Runway length (unfactored): 3,718 ft Runway elevation: 4,000 ft Weather: assume ISA conditionsRunway: Paved, level and dryHeadwind Component: 4 kta. approximately : 4,000 lbs.b. approximately : 3,500 lbsc. approximately : 4,500 lbs.d. approximately : 3,600 lbs.350Chapter 12Multi Engine Class BLanding5. At an aerodrome, the Landing Distance Available is 3,700 feet. For a multi-engine Class B aircraft, what must be the actual landing distance in order to comply with the landing regulations?a. 5,291 ftb. 2,587 ftc. 2,249 ftd. 3,700 ft6. By what factor must the Landing Distance Available for a multi engine class B aeroplane be divided by in order to fnd the maximum allowable landing distance?

and 35 ft is 1747 metres. This is then multiplied by 1.15. Multiplying 1747 metres by 1.15 makes the total distance 2009 metres. One power unit inoperative (dry runway). The horizontal distance from the brakes release point (BRP) to a point equidistant between VLOF and the point at which the aeroplane reaches 35 ft with the critical power unit inoperative. As an example, let us assume that this distance is 1950 metres. One power unit inoperative (wet runway). The horizontal distance from the brake release point (BRP) to the point at which the aeroplane is 15ft above the take-of surface, achieved in a manner consistent with the atainment of V2 by 35ft, assuming the critical power unit inoperative at VEF. Lastly, as an example, let us assume that this distance is 2001 metresOnce these three distances have been calculated by the manufacturer, the greatest of the three is then published as the certifed net take-of run required. In our example, the net take-of run required is 2009 metres. In the exam you will be presented with various distances and you must be able to work out which of the distances is selected as the net take-of run requiredNET ACCELERATE STOP DISTANCE REQUIREDThe accelerate-stop distance on a wet runway is the greatest of: All engines operating. The sum of the distances required to accelerate from BRP to the highest speed reached during the rejected take-of, assuming the pilot takes the frst action to reject the take-of at the V1 for take-of from a wet runway and to decelerate to a full stop on a wet hard surface, plus a distance equivalent to 2 seconds at the V1 for take-of from a wet runway. One engine inoperative. The sum of the distances required to accelerate from BRP to the highest speed reached during the rejected take-of, assuming the critical engine fails at VEF and the pilot takes the frst action to reject the take-of at the V1 for take-of from a wet runway with all engines operating and to decelerate to a full stop on a wet hard surface with one engine inoperative, plus a distance equivalent to 2 seconds at the V1 for take-of from a wet runway. The accelerate-stop distance on a dry runway.357Chapter 13Class A AircraftTake OffNET TAKE-OFF DISTANCE REQUIREDThe take-of distance required is the greatest of the following three distances: All engines operating. The horizontal distance travelled, with all engines operating, to reach a screen height of 35 ft multiplied by 1.15 One engine inoperative (dry runway). The horizontal distance from BRP to the point at which the aeroplane atains 35 ft, assuming the critical power unit fails at VEF on a dry, hard surface. One engine inoperative (wet runway). The horizontal distance from BRP to the point at which the aeroplane atains 15 ft, assuming the critical power unit fails at VEF on a wet or contaminated hard surface, achieved in a manner consistent with the achievement of V2 by 35ft.THE V SPEEDSVEF

The calibrated airspeed at which the critical engine is assumed to fail. It is used for the purpose of performance calculations. It is never less than VMCG. The speed VEF is a rather strange one. As per the certifcation specifcation defnition, VEF means the speed at which the critical engine is assumed to fail during take-of. VEF is selected by the aeroplane manufacture for purposes of certifcation testing, primarily to establish the range of speeds from which V1 may be selected and secondly to help determine the accelerate stop distance required. Lets us try and explain what VEF is all about.The defnition of V1 is the speed at which, if the failure of the critical engine was recognised, there is sufcient distance remaining to either abandon the take-of or continue the take-of. However, recognising that the engine has failed does take time, in fact its about 1 second. Therefore to recognise the engine failure at V1, the engine must have failed about 1 second before V1. The speed, at which the critical engine fails, so that it may be recognised at V1, is called VEF .V1 DECISION SPEEDThis is by far the most important speed in the take-of for Class A aeroplanes. V1 is called the decision speed. It is so called because V1 determines the outcome of a critical decision that must be made following an engine failure or other major critical systems failure. V1 is defned as being the maximum speed at which the pilot must take the frst action in order to stop the aeroplane within the remaining accelerate stop distance. V1 is also the minimum speed following engine failure that the pilot is able to continue the take-of within the remaining take-of distance.VGO is the lowest decision speed from which a continued take-of is possible within the take-of distance available. VSTOP is the highest decision speed from which the aeroplane can stop within the accelerate-stop distance available. These two speeds are the extremes of V1.358Chapter 13Class A AircraftTake OffThere are some rules about the speed for V1. These are shown in CAP 698 on page 2 of section 4 alongside the V1 defnition. It states that V1: may not be less than VEF plus the speed gained with the critical engine inoperative for the time between engine failure and the point at which the pilot applies the frst means of retardation. must not exceed VR must not exceed VMBE must not be less than VMCGIf the engine were to fail before V1, then the decision would be to abort the take-of. The reason is because, with only one engine operating, there would be insufcient take-of distance left to accelerate the aeroplane to the screen height. If the engine were to fail after V1, the decision is to continue the take-of. The reason is because the aeroplane is travelling too fast to be able to stop within the remaining accelerate stop distance available. In order to understand how V1 is derived, we need to consider a graph which is shown in fgure 13.1. This graph plots the take-of distance required and accelerate stop distance required based on a varying engine failure speed. Figure 13.1 A graph showing the ideal position of V1.Looking at the graph you can see that a V1 at the intersection point of the graph is the best V1 to use, simply because a V1 at that speed requires the least amount of feld required or the least amount of runway. The V1 at the intersection point of the curves is sometimes called the Idealised V1. It is also the V1 which makes the take-of distance required be the same length as the accelerate stop distance required, and therefore, this V1 speed is also called the Balanced V1 since it balances the required distances for the aeroplane. What this graph is also useful for is to see the efect of using higher or lower V1s. For example, if for whatever reason V1 was increased, notice that the accelerate stop distance required increases, the take of distance required decreases and the total feld required increases. However, should V1 be reduced from the intersection point, the accelerate stop distance decreases, the take-of distance increases and the total feld required increases. Trying to fgure these points out without the use of this simple graph would not be easy. So use this graph to help you understand the efect of increasing or decreasing V1.359Chapter 13Class A AircraftTake OffFACTORS AFFECTING V1It is important to detail what factors can infuence V1. In essence, whatever factors change either the accelerate stop distance required or the take-of distance required curves shown in fgure 13.1, will afect V1. However, there is simpler way to go through the list of factors afecting V1. In every aeroplane fight manual there will be a set of tables and or graphs which will allow you to calculate what V1 should be on any given day. On page 18 and 19 of section 4 in CAP 698 are such tables. Turn to page 18 and look at the second table from the top of the page where you see a column A and B etc. This table lists the three V speeds, including V1, against the aeroplane mass. This table can show us what the efect of certain factors are. MASS For example, lets look at what aeroplane mass does to V1. Look under column A and under V1. If the aeroplane mass was 50,000 kg, V1 would be 129 knots, but if the mass was increased to 65,000 kg then V1 would increase to 151 knots. Therefore, increasing mass increases V1. CONFIGURATION The next factor to afect V1 is the aeroplane confguration. Page 18 lists the V speeds for 5 degrees faps and page 19 lists the V speeds with 15 degrees of fap. Therefore comparing the speeds from these two pages, should tell us the efect of the faps. On page 18, which is 5 degrees fap and using column A again, a 55,000 kg mass requires a V1 speed of 137 knots, but on page 19, which is 15 degree fap and for the same mass of 55,000 kg, the V1 speed is now 130 knots. Therefore we can see that increasing the fap angle, decreases V1. DENSITY The next factor to afect V1 is density, but this is litle harder to observe than the previous two factors. On page 17 of section 4 is a small table, which has temperature on one axis and pressure altitude on the other. This is the density accountability table. Under low altitudes and low temperatures is band A. This is a high density band, Where as band F is situated at higher temperatures and higher altitudes which equates to much lower densities. We can now use these bands to see the efect density has on V1. Returning to page 18, notice that bands A, B, C, D, E and F are in the speed tables. Taking a mass of 50,000 kg in band A, which is high density, V1 is 129 knots, but in band B, C, D and E, the value of V1 is increasing. Therefore, as density falls, shown by going through bands A to E, V1 increases. SLOPE AND WIND The last two factors that afect V1 are runway slope and wind. Notice that at the top of page 18 and 19 there is a table which is titled Slope and Wind V1 adjustment. If there were a downslope of 2%, then with an aeroplane mass of 70,000 kg, V1 would have to be reduced by 3 knots. Whereas if the runway had an upslope of 2%, with the same mass, V1 must be increased by 4 knots. Therefore, downslope reduce V11 and upslope increase V1. The right hand side of the table is the correction for wind. So, for example, if there was a 15 knot tailwind, then with a mass of 70,000 kg, V1 would have to be reduced by 3 knots, whereas if there was a headwind of 40 knots for the same mass, V1 would have to increase by 1 knot. Therefore, tailwinds reduce V1, and headwinds increase V1.Having fnished discussing the main factors afecting V1, there are other infuences however, which may or may not change the value of V1. Two infuences on V1 in particular are the speed VMCG and VMBE. In CAP 698 on page 2 of section 4 we see the defnition of V1, but more importantly, at the end of the paragraph we read that V1 must not be less than VMCG, and not greater than VR and not greater than VMBE. Since V1 must between VMCG and VR and VMBE, then depending on the values of these speeds, they may or may not push V1 to be higher or lower than the ideal V1 speed. We will understand this beter once we have discussed what VMCG and VMBE are.360Chapter 13Class A AircraftTake OffVMCG - Ground Minimum Control SpeedVMCG is short for the ground minimum control speed, and it is described for you in CAP 698 at the botom of page 3 section 4. It states VMCG is the minimum speed on the ground at which the take-of can be safely continued, when the critical engine suddenly becomes in-operative with the remaining engines at take-of thrust. Lets try and understand what this actually means.When an engine fails, the remaining engine still generates thrust and this causes the aeroplane to yaw away from the live engine. The amount of yaw is a function of the amount of thrust the live engine is generating. Greater thrust from the live engine would generate more yaw. The only way to counteract this is to use the ailerons and the rudder to try and steer the aeroplane in the right direction. However, when the aeroplane is on the ground, you cannot use the ailerons to control the yaw otherwise you might bank the wing into the ground. Therefore the only available aerodynamic surface left to control the asymmetric yaw is the rudder. However, for the rudder to be efective enough at controlling the yaw, there must be sufcient airfow over it to ensure it has the required aerodynamic force. This minimum airfow speed over the rudder is VMCG. If the engine were to fail below this speed, then there is insufcient fow over the rudder to counteract the asymmetric yaw and therefore it is not possible to continue the take-of.The only factor that controls the value of VMCG is thrust, and since take-of thrust is more or less constant, then the only variable on the amount of take-of thrust generated is air density. The higher the air density, the more thrust that can be generated and therefore the more yaw that is generated when the engine fails, therefore the airfow over the rudder must be faster to make the rudder efective enough to counteract the yaw.The efect of air density on VMCG can be seen by looking at the second table from the botom on page 18 and 19 of section 4 in CAP 698. This table shows the variable of temperature on one side and the variable of pressure altitude on the other. Look at the VMCG in the table and notice that at low temperatures and low pressure altitudes where the air density would be high, the value of VMCG is also high. Therefore we can say that as density increases, VMCG increases. Although VMCA is the minimum control speed in the air, the factors that afect VMCA can for the purpose of the exam, are the same as for VMCG.VMCA / VMC - Air Minimum Control SpeedThe air minimum control speed. The minimum fight speed at which the aeroplane is controllable, with a maximum of 5 bank, when the critical engine suddenly becomes inoperative with the remaining engines at take-of thrust.VMCL - Landing Minimum Control SpeedThe minimum control speed during landing approach. The minimum speed with a wing engine inoperative where it is possible to decrease thrust to idle or increase thrust to maximum take-of without encountering dangerous fight characteristics.VMBE - Maximum Brake Energy Speed We stated that there were two particular speeds that can infuence V1. One of them was VMCG

which we just discussed the other was VMBE. Turning back to the top of page 3 of section 4 of CAP 698 you can see the description of VMBE. VMBE is the maximum brake energy speed and it represents the maximum speed on the ground from which an aeroplane can safely stop within the energy capabilities of the brakes. Essentially this means that if the take-of was abandoned at a speed higher than VMBE, and maximum braking force was applied, the brakes would not be able to safely bring the aeroplane to a stop regardless of how much runway was left. The brakes would most probably catch fre, melt and or disintegrate. 361Chapter 13Class A AircraftTake OffYou do need to be aware of the factors that control VMBE, but luckily, most manuals, and indeed CAP 698 has a VMBE graph or table with all the variables and factors on it that can afect VMBE. The graph concerned is on page 15 of section 4. If you need to see the efect of a variable, for example, mass, simply work through the graph but use two diferent masses. In this case the heavier mass has reduced VMBE. The variables that afect VMBE are pressure altitude, ambient air temperature, mass, slope and wind. Carefully examine each of these factors so you can see for yourself how they change VMBE. Remember, CAP 698 is for use in the exam, so if there are any questions which relate to VMBE, you can rest assured that a lot of information on VMBE is already in front of you.VMCG - V1 VMBEHaving looked at VMBE and VMCG we are now beter placed to understand why these two speeds play a role in infuencing V1. According to the rule, V1 must not be less than VMCG, as shown in fgure 13.2Figure 13.2 The relationship of V1 with VMCG and VMBE

V1 cannot be allowed to be less than VMCG because engine failure below VMCG means the aeroplane is uncontrollable and the defnition on V1 is that the take-of can be continued following engine failure.The rule also stated that V1 must not be greater than VMBE. Again, this makes sense, because at V1 the aeroplane must be able to stop or continue the take-of, but above the VMBE it is impossible to bring the aeroplane safely to a stop.Let us look at a scenario, where due to high density; the value of VMCG is higher than the idealised V1. In this case, take-of is prohibited. However, this problem is solvable. The chosen V1 can simply be increased until it is equal to or more than VMCG. However, notice that the accelerate stop distance increases, the take-of distance decreases and more importantly, the total feld length required increases. So long as the runway is as long as the total feld required, then moving V1 to this point is not a problem. Hopefully, by understanding this graph you are able to see the consequences to the required distance should V1 need to be moved from its ideal or balanced position due to pressure from VMCG and VMBE362Chapter 13Class A AircraftTake OffVMU - Minimum Un-stick Speed The speed VMU is defned as the minimum un-stick speed. VMU is slowest calibrated airspeed, at which, the aeroplane can safely lift of the ground, and continue the take-of. However, despite VMU being the lowest speed the aeroplane can safely lift of the runway, in actual operating conditions, the aeroplane does not lift of at this speed. The aeroplane is fown so that it actually lifts-of at a slightly faster speed. The reason is because VMU is very close to the stall speed, the aeroplane controllability is very sloppy, and lastly, in order to actually lift of at VMU some fairly dramatic actions take place which may be uncomfortable for the passengers.It may seem strange, but the aeroplane is actually able to lift at a speed where lift is less than weight. The reason being because, so long as the nose can be raised to high enough atitude, then there is a vertical component of thrust which, together with lift, balances weight. The amount of this vertical thrust is controlled in part by the amount of thrust generated, but also by the amount of nose up atitude the aeroplane can atain. This nose up atitude may be limited by the power of the elevator to push the tail plane down, or by the tail plane striking the runway in what is described as a tail strike. Hopefully now you are able to realise why it is unwise in operational conditions to lift the aeroplane of the ground at VMU. The actual speed the aeroplane will lift of, in operational fights, is called VLOF and we will discuss this speed next.VLOF - Lift-of Speed VLOF means the lift-of speed. VLOF is the calibrated airspeed at which the aeroplane frst becomes airborne which is at the moment when the main wheels have left the runway. VLOF should be faster than the minimum unstuck speed VMU. The margin above VMU is determined by several factors. For example, VLOF must not be less than 110% of VMU in the all engines- operating condition and 105% of VMU in the one engine inoperative condition. However, if the atitude of aeroplane in obtaining VMU was limited by the geometry of the aeroplane (i.e., tail contact with the runway), VLOF must not be less than 108% of VMU in the all-engines operating condition and 104% of VMU

in the one engine inoperative condition.VR - Rotation SpeedRotation speed, VR, is the speed at which the pilot initiates action to raise the nose gear of the ground, with the intention of becoming airborne. The pilot action is to pull back on the control column. This action defects the horizontal stabilizer to create a downward aerodynamic force. This force rotates the aeroplane about its lateral axis and will raise the nose wheel of the ground.VR may not be less than V1 1.05 VMC a speed such that V2 may be atained before 35ft. a speed such that if the aeroplane is rotated at its maximum practicable rate will result in a VLOF of not less than 1.1 VMU (all engines operating) or 1.05VMU (engine inoperative) [ if the aeroplane is geometry limited or elevator power limited these margins are 1.08 VMU (all engines) and 1.04 VMU (engine inoperative)]363Chapter 13Class A AircraftTake OffFigure 13.7Figure 13.3Lastly, as with other speeds, we need to examine the factors that afect VR. However, this is made easy for you. Similarly with V1, all the factors that can afect VR can be found by examining page 18 and 19 of section 4 in CAP 698. On page 18 and 19 examine the second table from the top. This table lists the three V speeds, including VR, against the aeroplane mass. This table can show us what the efect of certain factors is. For example, lets look at what aeroplane mass does to VR. If the aeroplane mass was 50,000 kg, VR would be 131 knots under density column A, but if the mass was increased to 65,000 kg then V1 would increase to 155 knots. Therefore, increasing mass increases VR. The next factor to afect VR is the aeroplane confguration. Page 18 lists the V speeds for 5 degrees faps and page 19 lists the V speeds with 15 degrees of fap. Therefore comparing the speeds from these two pages, should tell us the efect of the faps. On page 18, which is 5 degrees fap, a 55,000 kg mass requires a VR speed of 139 knots, but on page 19, which is 15 degree fap and for the same mass of 55,000 kg, the VR speed is now 131 knots. Therefore we can see that increasing the fap angle, decreases VR. The next factor to afect VR is density. On page 17 of section 4 is the density graph, You may recall that band A is a high density band, whereas band F equates to lower density. We can now use these bands to see the efect density has on VR. Returning to page 18, Taking a mass of 50,000 kg in band A, which is high density, VR is 131 knots, but in band B, C, D and E, the value of VR is increasing. Therefore we can state that as density decrease, VR increases. In modern airliners, VR is calculated not by looking at speed tables, but it will computed by the aeroplane once the relevant data is inserted in the fight management computer or multipurpose computer display unit. Once V1 and VR have been calculated by the pilots, they can be entered into the Flight Management System and thereafter shown to the pilots in the speed scale on the left hand side of the Primary Flight Display (PFD) or Electronic Atitude Director Indicator (EADI).364Chapter 13Class A AircraftTake OffEfects of early and over-rotationIf the aircraft is rotated to the correct atitude but at too low a speed, lift of will not occur until the normal VLOF, but there will be higher drag during the increased time in the rotated atitude, giving increased distance to lift of. Rotation to an atitude greater than the normal lift of atitude could bring the wing close to its ground stalling angle. Ground stall should not be possible with leading edge devices correctly set, so it is of extreme importance that these devices are set to the take-of position.JAR 25.107 requires: the take-of distance using a rotation speed of 5 knots less than VR shall not exceed the take-of distance using the established VR reasonable variations in procedures such as over rotation and out of trim conditions must not result in marked increases in take-of distance.Note: The expression marked increase in the take-of distance is defned as any amount in excess of 1% of the scheduled distance. (ACJ No.2 to JAR 25.107e4)V2MIN:The minimum take-of safety speed, with the critical engine inoperative.V2MIN May not be less than: 1.13 VSR for 2 and 3 engine turboprops and all turbojets without provision for obtaining a signifcant reduction in the one engine inoperative power-on stalling speed OR 1.08 VSR for turboprops with more than 3 engines and turbojets with provision for obtaining a signifcant reduction in the one engine inoperative power-on stalling speed. 1.1 VMCFigure 13.8Figure 13.4365Chapter 13Class A AircraftTake OffV2 - Take-Of Safety SpeedThe speed V2 is called the take-of safety speed. On page 3 of section 4 of CAP 698 it states that V2 is the target speed to be atained with one engine inoperative. In other words, V2 must be reached at or prior to the screen height. Why is V2 called the take-of safety speed, what is safe about reaching it? There are two main speeds which when fying close to, may be unsafe. The frst of these is stall speed and the second is the minimum control speed. Therefore, in order for V2 to be called a safe speed it must be a faster than these speeds. There is another reason why V2 is called the take-of safety speed. In the event of engine failure, V2 must be fown until the aeroplane reaches 400 ft. Therefore, the other safe feature about V2 is that the aeroplane is able to achieve a positive climb. In fact, V2 is the slowest speed which will enable the aeroplane to have sufcient excess thrust to climb above the minimum acceptable climb gradients.V2 may not be less than: V2MIN VR plus the speed increment atained up to 35 ft.Figure 13.9.Figure 13.5To analyse all the factors that can afect V2 turn to page 18 and 19 of section 4 in CAP 698. You will recall from a similar discussion on V1 and VR that these pages can show the efect of mass, confguration and density on V2. So ensure you can use these pages to see for yourself how these factors change V2.Once V2 is calculated by the pilots it can be entered into the fight management computer just like V1 and VR were. Having done this V2 will be displayed to the pilots in the speed scale on the left hand side of the Primary Flight Display or Electronic Atitude Director Indicator.V3The steady initial climb speed with all engines operating.366Chapter 13Class A AircraftTake OffPRESENTATION OF DATAA complete analysis of take-of performance requires account to be taken of any stopway and clearway available. As this is time consuming and will often give a maximum permissible take- of mass in excess of that required, simplifed data is often presented to permit a rapid assessment of the take-of mass. One method of doing this is to use balance feld data.BALANCED FIELDA balanced feld exists if the Take-of Distance is equal to the Accelerate-stop Distance. An aerodrome which has no stopway or clearway has a balanced feld. For an aeroplane taking of, if an engine failure occurs, the later the engine fails, the greater will be the accelerate-stop distance required but the less will be the take-of distance required At some speed the two distances will be equal. Figure 13.6 shows the variation of these distances graphically.Figure 13.10Figure 13.6The distance, point A in Figure 13.6 is the balanced feld length required for the prevailing conditions. It represents the maximum distance required for those conditions, because at whatever speed the engine fails, the distance is adequate, either to stop, if the failure occurs before V1 or to complete the take-of if the failure occurs after V1.UNBALANCED FIELD.For a given weight and conditions, the balanced feld V1 will give the optimum performance, since the TODR and the ASDR are equal. In some circumstances however this V1 will not be acceptable, as V1 must lie within the limits of VMCG , VR and VMBE . The following situations will give an unbalanced feld: V1 less than VMCG367Chapter 13Class A AircraftTake OffAt low weights and altitudes V1 for the balanced feld may be less than VMCG. In this case V1 would have to be increased to VMCG and so the TODR would be less, and the ASDR would be greater than the balanced feld length. The feld length required would be equal to the ASDR at VMCGFigure 13.11 Figure 13.7 V1 greater than VMBEAt high weight, altitude and temperature, the balanced feld V1 may exceed the VMBE . V1 would have to be reduced to VMBE giving a TODR greater, and an ASDR which is less, than the balanced feld length. The feld length required would be equal to the TODR at VMBE.368Chapter 13Class A AircraftTake OffFigure 13.12Figure 13.8 V1 greater than VRFor aircraft with good braking capabilities, the stopping distance will be short, giving a high balanced feld V1 speed. If this exceeds VR for the weight, V1 will have to be reduced to VR and the feld length required will be equal to the TODR at VR .Figure 13.13.Figure 13.9369Chapter 13Class A AircraftTake OffV1 RANGEIf the balanced feld available is greater than the balanced feld required for the required take-of mass and conditions there will be a range of speed within which V1 can be chosen. This situation is illustrated in Figure 13.10.Figure 13.14. Figure 13.10VGO is the frst speed at which the take-of can be completed within the distance available, and VSTOP is the last speed at which the accelerate-stop could be completed within the distance. The V1 speed can therefore be chosen anywhere between VGO and VSTOP.TAKE-OFF FROM AN UNBALANCED FIELDIf the take-of aerodrome is not a balanced feld, the balanced feld data can be used by assuming a balanced feld equal to the lesser of the Take-of Distance Available and the Accelerate-stop Distance Available. This distance may exceed the Take-of Run Available unless the TORA becomes limiting. The take-of mass obtained will of course be less than that which could have been obtained by taking account of stopway and clearway, but if the mass is sufcient for the fight, it will not be necessary to go into a more detailed analysis.370Chapter 13Class A AircraftTake OffFIELD LIMIT BRAKE RELEASE MASS / FIELD LIMIT MASSThis section of the chapter will focus on calculating the various limiting masses for take-of. Class A aeroplanes have data presented to the pilot in a diferent way than smaller class B aeroplanes. Whereas Class B aeroplane data for take-of would show what length of runway would be used for any given mass, Class A aeroplane data shows what maximum mass could be taken for a given runway length. This makes sense since Class A aeroplanes are used commercially and the interest of the airlines is to carry the maximum payload possible for the fight. Therefore most performance graphs or tables will give a mass as their outcome. The frst of these performance masses is the Field Limit brake release Mass. The feld limit brake release mass is the maximum mass that will allow the aeroplane to meet its feld length requirements at the airfeld concerned. Therefore, to be heavier than the feld limit mass would mean that the either the one engine inoperative or the all engine operative take-of run, take-of distance or accelerate stop distance exceeds the available distance at the airfeld. If you remember, airfelds can have diferent length of take-of run, take-of distance and accelerate stop distance available. Therefore, there should be many mass graphs. There should be mass graphs for ensuring that the mass is such that the take-of run required is within the take-of run available, that the take-of distance required is within the take-of distance available and lastly another mass to ensure the accelerate stop distance required is within the accelerate stop distance available. However, there is only one graph and only one assumed available distance to calculate the feld limit mass. The reason is for simplicity. The graph assumes that the take-of run available, the take-of distance available and the accelerate stop distance available are the same length even though the take-of distance available and accelerate stop distance available maybe longer. Therefore no stopways or clearways are accounted for. When the take-of distance available and the accelerate stop distance available are the same, the feld is described as being balanced. In this case the balanced feld length also happens to be the same length as the take-of run available because there are no stopways or clearways are assumed. To use the graph, make sure you only enter the take-of run available as the length of feld available.An example of a typical balanced feld length graph is shown in fgure 13.11. This graph is the exactly like the one shown in CAP 698 on page 9 of section 4. Notice at the botom of the graph there is only one feld distance to enter the graph with but of course an airfeld has many distances, such as the TODA and the ASDA. Because there is only one distance to enter into the graph it must be balanced feld length. The introduction to the graph is at the top of page 7 of section 4 in CAP 698 and it reiterates that the graph assumes a balanced feld. For unbalanced felds use the information under paragraph 2.5.1 on page 16 of section 4. This later information is for adjusting V1 when the feld is unbalanced.371Chapter 13Class A AircraftTake OffFigure 13.11 Field Limit Mass372Chapter 13Class A AircraftTake OffCLIMB GRADIENT LIMIT MASSThe feld limit mass is not the only mass that must be considered in the take-of, there are several more. The next mass to consider is the Climb Limit break release mass. The graph for calculating this mass is fgure 4.5 which is on page 11 of section 4 in CAP 698. The Climb limit mass is sometimes referred to as the Weight Altitude Limit or Mass Altitude Temperature limits, abbreviated to the WAT or MAT limit. Before we work though the climb limit mass graph lets try and understand what this mass means. The climb limit mass is the maximum mass that will enable the aeroplane to achieve a certain minimum climb performance. This minimum climb performance is the most severe of the climb gradient requirements. The most severe climb gradient requirement is in fact is 2.4% which will be covered later. In other words, if the mass of aeroplane was greater than the climb limit mass then the aeroplane may still be able to climb, but it will not achieve the minimum air gradients that the authorities have laid down. In other word the aeroplane would not achieve the climb requirements. The gradients for the climb requirements will be discussed in the next chapter but remember that these gradients are air gradients and are therefore unafected by wind. Figure 13.12 shows a typical presentation of the climb limited take-of mass and in found in CAP 698 on page 11 of section 4.373Chapter 13Class A AircraftTake OffFigure 13.12 Take-of Climb Limit / Climb Limit Mass374Chapter 13Class A AircraftTake OffTYRE SPEED LIMIT MASSThe reason for a tire speed limit is because naturally there is resistance between the wheel and the runway. As the wheel rotates this resistance generates heat. The greater the wheel speed and or the greater the load on the wheel, the greater the heat generated. Too much heat will not only disintegrate the tyre but it may also expand the air within the tire and may over pressurise it. This is dangerous and may result in a tire blow out, although there are fusible plugs in modern tires to help prevent this. As you can now understand, there is a maximum ground speed and maximum mass that the wheels can be subject to. The maximum ground speed that the tyre will experience will be at VLOF, and as a result, tyre speed limits are designed to be greater than or equal to the fastest VLOF. For most medium range jets the maximum tire speed limit is set at 195 knots which is about 225 miles per hour. Figure 13.13 shows a typical presentation of the tyre speed limited take-of mass and is found in CAP 698 on page 13 of section 4.375Chapter 13Class A AircraftTake OffFigure 13.13 Tyre Speed Limit Mass376Chapter 13Class A AircraftTake OffBRAKE ENERGY LIMITFor an aircraft of mass M, travelling at a true speed of V, the kinetic energy is MV2. If the aircraft is braked to a stop from this speed, a large proportion of this energy will go into the brakes as heat. The energy capacity of the brakes is limited and so for a given mass there will be a limiting speed from which a stop can be made. This will be a True Ground Speed, and so the corresponding IAS will vary with altitude, temperature and wind. Runway slope will also afect the speed, as a change in height involves a change in potential energy.The brake energy limit speed VMBE must not be less than the V1 speed. If it is, the mass must be reduced until V1 and VMBE are the same. The Flight Manual will give the amount of weight to be deducted for each knot that V1 exceeds VMBE. For most aircraft, VMBE will only be limiting in extremely adverse conditions of altitude, temperature wind and runway slope. In fact if you look at Figure 13.14 you will notice a grey area in the graph on the top left hand side. If the mass and pressure altitude falls within this grey area then VMBE will not be limiting. The same graph can be seen in CAP 698 on page 15 of section 4.377Chapter 13Class A AircraftTake OffFigure 13.14 Brake Energy Limit378Chapter 13Class A AircraftTake OffBRAKE COOLINGThe value of VMBE obtained from the data assumes that the brakes are at ambient temperature before the start of take-of. If a take-of is abandoned following a recent landing, or after prolonged taxying the brakes will already be at a fairly high temperature, and their ability to absorb further energy will be reduced. Data is given in the Manual to show the time to be allowed for the brakes to cool. AN example of a brake cooling graph is shown in fgure 13.15 and it can also be found in CAP 698 on page 50 of section 4.379Chapter 13Class A AircraftTake OffFigure 13.15 Shows a typical brake cooling schedule380Chapter 13Class A AircraftTake OffRUNWAY STRENGTHThe operating mass of the aircraft may be limited by runway strength considerations. The bearing strength of a pavement is expressed by a PCN (Pavement Classifcation Number) and this is compared to the ACN (Aircraft Classifcation Number). The UK system of classifcation is the LCN (Load Classifcation Number) but this can be converted into the PCN system. The PCN is compared to the ACN. Operation on the pavement is permissible if the ACN is less than or equal to the PCN. Because the PCN includes a safety factor, a 10% increase of ACN over PCN is generally acceptable for pavements that are in good condition and occasional use by aircraft with ACNs up to 50% greater than the PCN may be permited. In such circumstances the movement of the aircraft must be very closely monitored for damage to the aeroplane and pavement.MAXIMUM TAKE-OFF MASSConsideration of the mass determined by the feld length available, the climb requirement, the tyre speed limit, and the brake energy limit will determine the maximum mass for take-of. It will be the lowest of the masses given by the above limitations. This mass is called the Performance Limited Mass. The Performance Limited Mass must then be compared to maximum structural mass and the lower of the two masses is then selected as the take-of mass. This mass is known as the Regulated Take-of Mass. If there are obstacles to be considered on the take-of fight path this may determine a further limitation on take-of mass. Analysis of obstacle clearance limited mass is examined in Chapter 15.CALCULATING TAKE OFF SPEEDS AND THRUST SETTINGSWhen the maximum permissible take-of mass (regulated take-of mass) has been determined, it is necessary to fnd the corresponding take-of speeds and thrust setings. CAP 698 on page 17, 18, 19 and 20 of section 4 show the presentation of the take-of speeds V1, VR and V2 and the % N1 for take of. TAKE-OFF SPEEDSHaving chosen the regulated take-of mass, which for the purpose of an example we shall assume is 57900 kg, we are now able to select the take-of V speeds of V1, VR and V2. Before we calculate these speeds the speed band needs to be selected. At the botom of page 17 of section 4 of CAP 698 is a small table and is reproduce in fgure 13.16. This graph is the density correction graph for the take-of V speeds. As an example, let us assume a temperature of 25 degrees Celsius at an aerodrome pressure altitude of 2,000 ft. In our example, the speed band to use is speed band B.381Chapter 13Class A AircraftTake OffFigure 13.16 The density band selection for the take-of V speeds.Turning over the page in CAP 698 from page 17 of section 4 to page 18 and 19 you can see the speed tables. There are two sets of speed tables. The one on page 18 is for a 5 degrees fap seting and page 19 is for a 15 degree fap seting. For our example we will assume 5 degree faps with a regulated take-of mass of 57,900 kg. Using the speed tables for a 5 degree fap (page 18 section 4) identify the column for speed band B (should be roughly in the middle of the page). This is reproduced for you in fgure 13.17 and the areas to concentrate on are highlighted in red for you.Figure 13.17The left hand scale of the speed table is mass. Notice that our example regulated take-of mass of 57,900 kg lies between the 55,000kg and 60,000 kg marks. Therefore interpolation must be used when working out our V speeds. V1 for 55,000 kg is 138 knots and for 60,000 kg V1 is 145. Correct interpolation for 57,900 kg would make V1 equal to 142.06 knots which should be rounded but we will do this shortly. Carrying out the same exercise for VR makes VR for 57,900 kg equal to 145 knots and V2 equal to 152 knots. However, V1 must be corrected for slope and wind as shown in the table at the top of the page 18 and again is reproduced for you in fgure 13.18382Chapter 13Class A AircraftTake OffFigure 13.18 Slope and Wind adjustment for V1For a 2% upslope and a mass of 57,900 kg, interpolation shows that V1 must be increased by 1.79 knots. For a 20 knot headwind, V1 must be increased by 0.5 of a knot. Therefore the total correction to V1 is to increase it by 2.29 knots. Adding 2.29 knots to the original V1 of 142.06 that we calculated earlier makes V1 to be 144.35 knots, which is rounded to 144 knots. Having fnished with V1, VR and V2, there are two other speeds to check just though. The frst is VMCG.Looking further down the page you can see the table for calculating VMCG which is reproduced for you in fgure 13.19Figure 13.19 The VMCG tableUsing 25C and 2000 ft as our pressure altitude would make VMCG to be 112 knots. Remember from our theory that V1 must not be less than VMCG which in our example it isnt.Lastly the speed VMBE also needs calculating. The graph for this is on page 15 of section 4 of CAP 698. We have already described this graph and how to use it. Using fgure 13.14, if we use our example airfeld conditions of 2000 ft pressure altitude, 25C and our regulated take-of mass of 57,900 kg then the VMBE is 175 knots. However, there are some corrections to make. In our example we had a 2% upslope, this means we need to increase VMBE by 4 knots. There is also a wind correction to be made. In our example we assume a 20 knot headwind and this means VMBE must increase by 6 knots. The total correction would increase VMBE to 185 knots. 383Chapter 13Class A AircraftTake OffAll the relevant take-of V speeds have now been calculated based upon our regulated take-of mass of 57,900 kg.CORRECTION FOR STOPWAY AND CLEARWAYThe speeds shown in the tables that we have just used are based on a balanced feld length (TORA = TODA= ASDA) and are not valid if the take-of mass has been derived using stopway or clearway. Where this is the case the V1 may be adjusted for the efects of stopway or clearway from the table below.Figure 13.20384Chapter 13Class A AircraftTake OffTHRUST SETTING (% N1)The thrust seting values as shown in CAP 698 on page 20, 21, 22 and 23 of section 4. Use the tables on these pages to select the appropriate thrust seting for take-of and for the climb using the conditions at the airfeld.STABILISER TRIM SETTINGThe stabiliser trim seting appropriate to the C.G. position and take-of mass can be read from the table below and are shown in CAP 698 at the botom of pages 18 and 19 of section 4.Figure 13.21385Chapter 13Class A AircraftTake OffQUESTIONS1. For a given take-of mass, the maximum brake energy limit speed (VMBE), as an indicated airspeed, will:a. Decrease with increasing altitude, and decrease with increasing temperature.b. Increase with increasing altitude and increase with increasing temperature.c. Decrease with increasing altitude, and increase with increasing temperature.d. Not change with altitude, but decrease with increasing temperature.2. Provided all other parameters stay constant. Which of the following statements will decrease the take-of ground run?a. Decreased take-of mass, increased pressure altitude, increased temperature.b. Decreased take-of mass, increased density, increased fap seting.c. Increased pressure altitude, increased outside air temperature, increased take-ofmass.d. Increased outside air temperature, decreased pressure altitude, decreased fap seting.3. A multi engine aeroplane is fying at the minimum control speed (VMCA). Which parameter(s) must be maintainable after engine failure?a. Heading, altitude and a positive rate of climb of 100 ft/min.b. Altitude.c. Straight fight.d. Straight fight and altitude.4. How is VMCA infuenced by increasing pressure altitude?a. VMCA decreases with increasing pressure altitude.b. VMCA increases with pressure altitude higher than 4000 ft.c. VMCA increases with increasing pressure altitude.d. VMCA is not afected by pressure altitude.5. Which of the following speeds can be limited by the maximum tyre speed?

a. Lift-of groundspeed.b. Lift-of IAS.c. Lift-of TAS.d. Lift-of EAS.6. A higher outside air temperature (OAT):a. decreases the brake energy limited take-of mass.b. increases the feld length limited take-of mass.c. increases the climb limited take-of mass.d. decreases the take-of distance.386Chapter 13Class A AircraftTake Off7. The take-of performance requirements for Class A transport category aeroplanes are based upon:a. failure of critical engine.b. failure of critical engine or all engines operating which ever gives the largest take-ofdistance.c. all engines operating.d. only one engine operating.8. Maximum and minimum values of V1 can be limited by:a. VR and VMCGb. V2 and VMCAc. VR and VMCAd. V2 and VMCG9. During the certifcation fight testing of a twin engine turbojet aeroplane, the actual demonstrated take-of distances are equal to:1,547 m with all engines operating1,720 m with failure of the critical engine at V1 and with all other things remaining unchanged.The take-of distance adopted for the certifcation fle is:a. 1,547 m.b. 1,720 m.c. 1,779 m.d. 1,978 m.10. The minimum value that V2 must exceed air minimum control speed is by:a. 15%b. 20%c. 30%d. 10%11. With regard to a take-of from a wet runway, which of the following statements is correct?a. Screen height cannot be reduced.b. The screen height can be lowered to reduce the mass penalties.c. When the runway is wet, the V1 reduction is sufcient to maintain the same margins on the runway length.d. In case of a thrust reverser inoperative, the wet runway performance information can still be used.12. Balanced V1 is selected:a. for a runway length limited take-of with a clearway to give the highest mass.b. if it is equal to V2.c. if the accelerate stop distance required is equal to the one engine out take-of distance required.d. for a runway length limited take-of with a stopway to give the highest mass.387Chapter 13Class A AircraftTake Off13. How is V2 afected if take-of faps at 20 is chosen instead of take-of faps at 10?a. V2 increases in proportion to the angle at which the faps are set.b. V2 has no connection with take-of fap seting, as it is a function of runway length only.c. V2 decreases if not restricted by VMCA.d. V2 has the same value in both cases.14. Which statement regarding the infuence of a runway down-slope is correct for a balanced take-of? Down-slope:a. increases V1 and reduces the accelerate stop distance required (ASDR).b. reduces V1 and increases the accelerate stop distance required (ASDR).c. increases V1 and increases the take-of distance required (TODR).d. reduces V1 and reduces take-of distance required (TODR).15. Ignoring the minimum control speed limitation, the lowest take-of safety speed (V2 min) is:a. 1.15 Vs for all turbojet aeroplanes.b. 1.20 Vs for all turboprop powered aeroplanes.c. 1.13 VSR for two-engined and three-engined turbo-propeller powered aeroplanes;d. 1.13 VSR for turbo-propeller powered aeroplanes with more than three engines.16. During the fight preparation a pilot makes a mistake by selecting a V1 greater than that required. Which problem will occur when the engine fails at a speed immediately above the correct value of V1?a. The stop distance required will exceed the stop distance available.b. The one engine out take-of distance required may exceed the take-of distance available.c. V2 may be too high so that climb performance decreases.d. It may lead to over-rotation.17. The speed V2 of a jet aeroplane must be greater than: (assume the aeroplane has provisions for obtaining a signifcant reduction in the one-engine-inoperative power-on stall speed)a. 1.13 VMCG.b. 1.05 VLOF.c. 1.3 V1.d. 1.08 VSR.18. When an aircraft takes of with the mass limited by the TODA or feld length:a. the actual take-of mass equals the feld length limited takeof mass.b. the distance from brake release to V1 will be equal to the distance from V1 to the 35 feet point.c. the balanced take-of distance equals 115% of the all engine take-of distance.d. the end of the runway will be cleared by 35 feet following an engine failure at V1.388Chapter 13Class A AircraftTake Off19. A runway is contaminated by a 0.5 cm layer of wet snow. The take-of is nevertheless authorized by a light-twins fight manual. The take-of distance in relation to a dry runway will be:a. very signifcantly decreased.b. increased.c. unchanged.d. decreased.20. What will be the infuence on the aeroplane performance at higher pressure altitudes?a. It will increase the take-of distance.b. It will decrease the take-of distance.c. It will increase the take-of distance available.d. It will increase the accelerate stop distance available.21. During certifcation test fights for a turbojet aeroplane, the actual measured take-of runs from brake release to a point equidistant between the point at which VLOF is reached and the point at which the aeroplane is 35 feet above the take-of surface are:1,747 m, all engines operating1,950 m, with the critical engine failure recognized at V1, and all the other factors remaining unchanged. Considering both possibilities to determine the take-of run (TOR). What is the correct distance?a. 1,950 m.b. 2,009 m.c. 2,243 m.d. 2,096 m.22. Given that:VEF = Critical engine failure speed VMCG = Ground minimum control speed VMCA = Air minimum control speed VMU = Minimum un-stick speed V1 = Take-of decision speed VR = Rotation speed V2 min. = Minimum take-of safety speedThe correct formula is?a. 1.05 VMCA is less than or equal to VEF, VEF is less than or equal to V1b. 1.05 VMCG is less than VEF, VEF is less than or equal to VRc. V2min is less than or equal to VEF, VEF is less than or equal to VMUd. VMCG is less than or equal to VEF, VEF is less than V123. If the feld length limited take of mass has been calculated using a Balanced Field Length technique, the use of any additional clearway in take-of performance calculations may allow:a. a greater feld length limited take of mass but with a higher V1.b. the obstacle clearance limit to be increased with no efect on V1.c. the obstacle clearance limit to be increased with an higher V1.d. a greater feld length limited take of mass but with a lower V1.389Chapter 13Class A AircraftTake Off24. The result of a higher fap seting up to the optimum at take-of is:a. a higher V1.b. a longer take-of run.c. a shorter ground roll.d. an increased acceleration.25. For class A aeroplanes the take-of run is:a. the horizontal distance along the take-of path from the start of the take-of to a point equidistant between the point at which VLOF is reached and the point at which the aeroplane is 35 ft above the take-of surface.b. 1.5 times the distance from the point of brake release to a point equidistant between the point at which VLOF is reached and the point at which the aeroplane atains a height of 35 ft above the runway with all engines operative.c. 1.15 times the distance from the point of brake release to the point at which VLOF is reached assuming a failure of the critical engine at V1.d. the distance of the point of brake release to a point equidistant between the point at which VLOF is reached and the point at which the aeroplane atains a height of 50 ft above the runway assuming a failure of the critical engine at V1.26. Which statement is correct for a Class A aeroplane?a. VR must not be less than 1.05 VMCA and not less than 1.1 V1.b. VR must not be less than 1.05 VMCA and not less than V1.c. VR must not be less than VMCA and not less than 1.05 V1.d. VR must not be less than 1.1 VMCA and not less than V1.27. During certifcation fight testing on a four engine turbojet aeroplane the actual take-of distances measured are:2,555 m with all engines operating 3,050 m with failure of the critical engine recognised at V1 and all other things being equal.The take-of distance adopted for the certifcation fle is:a. 3,050 mb. 3,513 mc. 2,555 md. 2,938 m28. When the outside air temperature increases, then:a. the feld length limited take-of mass decreases but the climb limited take-of mass increases.b. the feld length limited take-of mass increases but the climb limited take-of mass decreases.c. the feld length limited take-of mass and the climb limited take-of mass decreases.d. the feld length limited take-of mass and the climb limited take-of mass increases.29. In case of an engine failure which is recognized at or above V1:a. the take-of should be rejected if the speed is still below VR.b. the take-of must be continued.c. the take-of must be rejected if the speed is still below VLOF.d. a height of 50 ft must be reached within the take-of distance.390Chapter 13Class A AircraftTake Off30. VR cannot be lower than:a. 105% of V1 and VMCA.b. 1.2 Vs for twin and three engine jet aeroplane.c. 1.15 Vs for turbo-prop with three or more engines.d. V1 and 105% of VMCA.31. If the performance limiting take-of mass of an aeroplane is brake energy limited, a higher uphill slope would:a. have no efect on the maximum mass for take-of.b. decrease the required take-of distance.c. increase the maximum mass for take-of.d. decrease the maximum mass for take-of.32. In which of the following distances can the length of a stopway be included?a. In the accelerate stop distance available.b. In the one-engine failure case, take-of distance.c. In the all-engine take-of distance.d. In the take-of run available.33. In the event of engine failure below V1, the frst action to be taken by the pilot in order to decelerate the aeroplane is to:a. apply wheel brakes.b. deploy airbrakes or spoilers.c. reduce the engine thrust.d. reverse engine thrust.34. The climb limited take-of mass can be increased by:a. selecting a lower VR.b. a lower fap seting for take-of and selecting a higher V2.c. selecting a lower V1.d. selecting a lower V2.35. Which is the correct sequence of speeds during take-of?a. V1, VR, V2, VMCA.b. VMCG, V1, VR, V2.c. V1, VMCG, VR, V2.d. V1, VR, VMCG, V2.36. Which of the following distances will increase if you increase V1?a. All Engine Take-of distance.b. Take-of run.c. Accelerate Stop Distance.d. Take-of distance.391Chapter 13Class A AircraftTake Off37. If the value of the balanced V1 is found to be lower than VMCG, which of the following is correct?a. The ASDR will become greater than the one engine out take-of distance.b. The take-of is not permited.c. The one engine out take-of distance will become greater than the ASDR.d. The VMCG will be lowered to V1.38. For this question use Figure 4.4 in CAP 698 Section 4.For an example twin engine turbojet aeroplane two take-of fap setings (5 and 15) are certifed.Given: Field length available = 2,400 m Outside air temperature = - 10C Airport pressure altitude = 7,000 ftThe maximum allowed take-of mass is:

a. 55,000 kg b. 70,000 kg c. 52,000 kg d. 56,000 kg392Chapter 13Class A AircraftTake OffANSWERS1. A2. B3. C4. A5. A6. A7. B8. A9. C10. D11. B12. C13. C14. D15. C16. A17. D18. A19. B20. A21. B22. D23. D24. C25. A26. B27. A28. C29. B30. D31. C32. A33. C34. B35. B36. C37. B38. D393Chapter 14Class AAdditional Take Off ProceduresCHAPTER FOURTEENCLASS A: ADDITIONAL TAKE OFF PROCEDURESContentsNON STANDARD TAKE-OFF PROCEDURES. . . . . . . . . . . . . . . . . . . . . . 395CONTAMINATED RUNWAYS. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395TAKE-OFF WITH INCREASED V2 SPEED . . . . . . . . . . . . . . . . . . . . . . . . 398TAKE-OFF WITH REDUCED THRUST . . . . . . . . . . . . . . . . . . . . . . . . . . 399TAKE-OFF WITH ANTI-SKID INOPERATIVE. . . . . . . . . . . . . . . . . . . . . . 401QUESTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402ANSWERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 404394Chapter 14Class AAdditional Take Off Procedures395Chapter 14Class AAdditional Take Off ProceduresNON STANDARD TAKE-OFF PROCEDURESThe procedure to determine the take-of mass, take-of speeds, and thrust setings for the normal take-of procedure is given in the previous chapter. Chapter 14 gives additional procedures to cover: take-of with contaminated runway take-of with increased V2 speed take-of with reduced thrust Take-of with anti-skid inoperativeHowever, most of these procedures can be found in CAP 698 on page 24 to 34 of section 4. Do not worry about having to remember too much about these procedures, CAP 698 adequately details and describes not only the theory of the procedures but also the methodology of each procedure.CONTAMINATED RUNWAYSThis procedure is detailed on page 24 of section 4 in CAP 698. A runway is considered to be contaminated when more than 25% of the runway surface area (whether in isolated areas or not) within the required length and width being used, is covered by surface water, more than 3mm. deep, or by slush or loose snow, equivalent to more than 3mm. of water.Slush, loose snow or standing water on the runway will afect both the Take-of distance Required and the Accelerate-stop Distance Required. The Take-of Distance Required will increase because of the additional wheel drag and impingement drag. The accelerate-stop distance will increase because of the increased distance to decelerate and the increased distance to stop resulting from the reduced runway coefcient of braking friction. For given distances available, the maximum take-of mass and the V1 will therefore be reduced compared to the dry runway. The greater the depth of contamination, the greater the mass reduction and the less the V1 reduction.The supplementary performance information required by JAR-OPS 1 should include accelerate-stop distance, take-of distance and take-of run appropriate to the relevant contaminant, derived in a similar manner to those distances with a wet runway.The acceleration distance should take account of the additional drag due to the gear displacement drag, and the spray impingement drag, and the decrease of drag which occurs above the aquaplaning speed. For rotating tyres or tyres going from a dry surface to a fooded surface, the hydroplaning speed (VP) is calculated using the formula below396Chapter 14Class AAdditional Take Off ProceduresThe hydroplaning speed for non rotating tyres or to fnd the speed below which the aquaplaning will stop is shown below.PROCEDUREUse either the fgures below or the tables shown in CAP 698 on page 25, 26 and 27 of section for the following description of the procedure to follow for contaminated runways.397Chapter 14Class AAdditional Take Off ProceduresFigure 14.1 Sample data for a runway with 2mm contamination398Chapter 14Class AAdditional Take Off ProceduresThe Determination of Take-Of Massa) Calculate the normal limiting take-of mass for a dry runway i.e., feld length limit, climb limit or obstacle limit.b) Select the table(s) appropriate to the depth of contaminant (interpolating if necessary).c) Enter the left column of the top table at the normal limiting take-of mass, travel right to the aerodrome pressure altitude column. Interpolate for mass and pressure altitude, if necessary. Extract the mass reduction. Calculate maximum take-of mass for a contaminated runway by subtracting the mass reduction from the normal performance limiting take-of mass.d) If in the shaded area, proceed to the botom table. Enter the left column with the take-of run available (TORA), move right to the appropriate aerodrome pressure altitude column. Interpolate as necessary. Extract the maximum permissible takeof mass. Make V1 = VMCG.e) The lower of the two values from c) and d) above is the maximum take-of mass for a contaminated runway.f) Calculate all the V speeds for the actual take-of mass as given in e). g) If not in the shaded area in c) above, then re-enter the left table at the actual mass to determine the V1 reduction to be made.h) Apply the reduction to V1. If adjusted V1 is less than VMCG, take-of is not permited.TAKE-OFF WITH INCREASED V2 SPEEDThis particular procedure is used when the performance limiting mass is the climb limit mass. In other words, the climb performance is poor and is severely restricting the potential mass of the aeroplane. Before we start, it is important to understand that in the event of engine failure, the initial climb out speed is V2. However, V2 is not the best climb angle speed. V2 is considerably slower than the best angle of climb speed, which, if you recall is called VX. As an example, for a typical new generation 737, VX is 80 knots faster than V2; therefore climbing out at V2 produces a climb angle much less than if the aeroplane were to climb out at VX.What the improved climb procedure aims to achieve is to increase V2 to be closer to VX. This will greatly enhance the climb performance. Understanding the concepts of the increased V2

procedure would be best illustrated by working through an example where the feld limit mass is 61,000 kg and the climb limit mass is 52,000 kg. The performance limited mass is always the lower mass and therefore the mass for take-of must be 52,000 kg. This is a shame, since the runway can allow a far greater mass. Therefore, taking of with only 52,000 kg would mean there would be a signifcant proportion of the runway left. This can provide a clue as to the solution.With all the excess of runway it would be possible to stay on the runway for longer during the take-of to build up more speed, this will ensure that at rotation and at the screen height, a faster V2 will be reached and this faster V2 will be much closer to VX. This ensures that the climb performance signifcantly improves. As a result of the improved climb performance, the climb limit mass can increase, which would increase the performance limited take-of mass and provide an improved regulated take-of mass.399Chapter 14Class AAdditional Take Off ProceduresThe information at the top of page 28 of section 4 in CAP 698 introduces the concept of the procedure very well, so use this introduction in the exam if you are unsure of what the procedure entails. Below this introduction is the detailed methodology of the procedure itself which is reproduced in the next paragraph.PROCEDUREUse the graphs shown in CAP 698 on page 29 and 30 of section for the following procedure for an increased V2 take-of. a) Select the set of graphs appropriate to the fap seting on the improved climb performance feld length limit graph (CAP 698 Figure 4.15).b) Enter the relevant left-hand graph with the value of the feld length limit mass minus the climb limit mass. Travel vertically up to the normal climb limit mass line.c) From this intersection move horizontally left to the vertical axis to read the climb mass improvement and horizontally right to the vertical axis to read the increase to apply to V1.d) Continue horizontally right to the reference line of the right-hand graph. From this point interpolate and follow the grid lines to reach a vertical input in the right-hand graph of the normal climb limit mass.e) From this intersection, travel horizontally right to the vertical axis to read the increase to apply to VR and V2.f) Repeat this process in the improved climb performance tyre speed limit graph (CAP 698 Figure 4.16) except that the initial entry point is the tyre limit mass minus the climb limit mass.g) The lower of the two mass increases is that which must be used together with its associated speed increases.h) Add the mass increase to the normal climb mass limit.i) Determine the V speeds for this increased mass.j) Apply the speed increases to the appropriate speeds. Check VMBE.TAKE-OFF WITH REDUCED THRUSTThe third additional type of procedure is probably the most common and it is referred to by many diferent names and is detailed on page 31 of section 4 in CAP 698. Use this page to help you, it describes all the relevant information you need for you. So again, dont worry about having to remember all the information about it. This third procedure is referred to as the reduced thrust take-of, variable thrust take-of or assumed temperature take-of. However, Airbus uses the term Flexible take-of. The main reason for doing this procedure is to preserve engine life and also to help reduce noise. The procedure can be used any time the actual take-of mass is less than the maximum permissible take-of mass and that there is an available distance that greatly exceeds that which is required. The maximum reduction in thrust from the full rated take-of thrust value is 25%. 400Chapter 14Class AAdditional Take Off ProceduresTake-of with reduced thrust is not permited with: icy or very slippery runways contaminated runways ant-skid inoperative reverse thrust inoperative increased V2 procedure PMC ofReduced thrust take of procedure is not recommended if potential windshear conditions exist.PROCEDUREEssentially this procedure involves pretending or assuming that the temperature is a lot hoter than it actually is. Imagine for the moment that the outside air temperature was continually increasing and as a result the thrust produced by the engines continually decreasing. There will eventually be a temperature beyond which there will be insufcient thrust to complete a take-of. This temperature is then used as the assumed temperature and the thrust equating to this temperature is then set as the take-of thrust.The procedure described below can also be found in CAP 698 on page 31 of section 4.It is frst necessary to determine the most limiting performance condition. The only common parameter to enable comparison is that of temperature. Thus the maximum permissible temperature must be calculated for the actual take-of mass from each of the following: feld limit graph climb limit graph tyre-speed limit graph obstacle limit graphFrom these temperatures, select the lowest and ensure that it does not exceed the environmental limit. If it does, then the environmental limit becomes the assumed temperature.a) Calculate the maximum assumed temperature from Figure 4.17 a or b, as appropriate. Enter the left column with the actual ambient temperature and read the maximum temperature in the column appropriate to the aerodrome pressure altitude.b) From Figure 4.17c on botom line, determine the minimum assumed temperature for the aerodrome pressure altitude.c) From the same table, for the assumed temperature to be used, determine the maximum take-of % N1. Add 1.0% N1 if air conditioning packs are of. The assumed temperature used must neither exceed the maximum from paragraph a) above nor be below the minimum from paragraph b) above.d) Enter the left column of Figure 4.17d with assumed temperature minus ambient temperature. Travel right along the line to the column appropriate to the ambient temperature, interpolating if necessary. Read the % N1 adjustment.e) Subtract the value determined at paragraph d) from that at paragraph c) to determine the % N1 to be set at take-of.401Chapter 14Class AAdditional Take Off ProceduresTAKE-OFF WITH ANTI-SKID INOPERATIVE(Simplifed Method)The last additional take-of procedure is that used when the anti skid system is inoperative. You may think that this is not important for take-of, but bear in mind, Class A aeroplanes have to demonstrate that in the event of engine failure, the aeroplane is able to stop within the confnes of the runway. Therefore the accelerate stop distance required must be less than or equal to the feld available. If the anti-skid system doesnt work, then the stopping ability will be severely reduced and will cause the accelerate stop distance to increase dramatically. To solve the problem, V1 is reduced. You may recall, that reducing V1 decrease the accelerate stop distance, but the side efect is that it increases the take-of distance (see fgure 13.1). To resolve this, the mass of the aeroplane is reduced, which will decreases both the accelerate stop distance and take-of distance required so that they remain within the available feld lengths.In summary then, with the anti-skid system inoperative, V1 and aeroplane mass must be decreased.The anti-skid inoperative procedure is detailed on page 34 of section 4 in CAP 698. At the top of that page is a very clear introduction to the procedure and below this is the method of calculating the mass and V1 reduction402Chapter 14Class AAdditional Take Off ProceduresQUESTIONS1. With regard to a take-of from a wet runway, which of the following statements is correct?a. Screen height cannot be reduced.b. The screen height can be lowered to reduce the mass penalties.c. When the runway is wet, the V1 reduction is sufcient to maintain the same margins on the runway length.d. In case of a thrust reverser inoperative, the wet runway performance information can still be used.2. For a take-of from a contaminated runway, which of the following statements is correct?a. Dry snow is not considered to afect the take-of performance.b. A slush covered runway must be cleared before take-of, even if the performance data for contaminated runway is available.c. The performance data for take-of is determined in general by means of calculation, only a few values are verifed by fight tests.d. The greater the depth of contamination at constant take-of mass, the more V1 has to be decreased to compensate for decreasing friction.3. Reduced take-of thrust is prohibited when:a. it is dark.b. the runway is wet.c. obstacles are present close to the end of the runway.d. the runway is contaminated.4. If the antiskid system is inoperative, which of the following statements is true?a. It has no efect on the accelerate stop distance.b. Take-of with antiskid inoperative is not permited.c. The accelerate stop distance increases.d. The accelerate stop distance decreases.5. A runway is contaminated by a 0,5 cm layer of wet snow. The take-of is nevertheless authorized by a light-twins fight manual. The take-of distance in relation to a dry runway will be:a. very signifcantly decreased.b. B increased.c. unchanged.d. decreased.6. Reduced take-of thrust is prohibited when:a. the runway is wet.b. the OAT is ISA +10C.c. anti skid is unserviceable.d. it is dark.403Chapter 14Class AAdditional Take Off Procedures7. Reduced take-of thrust should normally not be used when:a. it is dark.b. the runway is dry.c. the runway is wet.d. windshear is reported on the take-of path.8. When V1 has to be reduced because of a wet runway, the one engine out obstacle clearance / climb performance:a. increases / increases.b. remains constant / remains constant.c. decreases / decreases.d. decreases / remains constant.9. The climb limited take-of mass can be increased by:a. selecting a lower VR.b. a lower fap seting for take-of and selecting a higher V2.c. selecting a lower V1.d. selecting a lower V2.10. Due to a lot of standing water on the runway the feld length limited take-of mass will be:a. only higher for three and four engine aeroplanes.b. lower.c. higher.d. unafected.11. Reduced take-of thrust:a. can be used if the headwind component during take-of is at least 10 kt.b. has the beneft of improving engine life.c. can be used if the actual take-of mass is higher than the performance limited take-of mass.d. is not recommended at very low temperatures.12. Which statement about reduced thrust is correct?a. In case of reduced thrust V1 should be decreased.b. Reduced thrust can be used when the actual take-of mass is less than the feld length limited take-of mass.c. Reduced thrust is primarily a noise abatement procedure.d. Reduced thrust is used in order to save fuel.13. What is the efect of a greater contamination depth on the reduction to the take-of mass and the reduction to V1 respectively?a. Increase, decreaseb. Decrease, decreasec. Increase, increased. Decrease, increase404Chapter 14Class AAdditional Take Off ProceduresANSWERS1. B2. C3. D4. C5. B6. C7. D8. D9. B10. B11. B12. B13. A405Chapter 15Class ATake Off ClimbCHAPTER FIFTEENCLASS A: TAKE OFF CLIMBContentsTAKE-OFF CLIMB . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407SEGMENTS OF THE TAKE-OFF CLIMB . . . . . . . . . . . . . . . . . . . . . . . . . 407OBSTACLE CLEARANCE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409NOISE ABATEMENT PROCEDURES . . . . . . . . . . . . . . . . . . . . . . . . . . . 412QUESTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415ANSWERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 420406Chapter 15Class ATake Off Climb407Chapter 15Class ATake Off ClimbTAKE-OFF CLIMBThe take-of climb or take-of fight path extends from 35 ft above the take-of surface to 1500 ft above the take-of surface. However, with a contaminated runway take-of, the take-of climb begins at 15 ft and not 35 ft. The point on the ground directly below the 35 ft screen is called reference zero. There are two main requirements that must be met within the take-of climb and these requirements are based upon an engine failure occurring at VEF. Remember that performance of Class A aeroplanes must account for engine failure in all fight phases. The frst requirement is that the aeroplane must be able to achieve the minimum climb gradients and secondly the aeroplane must be able to maintain sufcient obstacle clearance. Remember that the climb gradient requirements are air based gradients and the obstacle clearance requirement use ground based gradients.When assessing compliance with the regulations, the manufacturer or operator may either use a continuous demonstrated take-of climb or a segmented take-of climb. Segmenting the take-of climb does make the requirements and the procedure a litle easier to comprehend; therefore we will use a segmented take-of climb profle as do most operators and manufacturers.SEGMENTS OF THE TAKE-OFF CLIMBThe take-of climb is generally split into four unique segments and these are shown in fgure 15.1. Each segment is characteristic of a distinct change in aeroplane confguration, speed and or thrust with various actions and climb gradient requirements. Generally there are four segments and you will need to learn what unique characteristics defne each segment.Figure 15.1 The segments of the take-of climb for typical Class A aeroplane.408Chapter 15Class ATake Off ClimbSEGMENT 1The take-of fight path starts once the take-of is complete, in other words at 35 ft with aeroplane at V2 with one engine inoperative. The 35 ft screen marks the start point of segment 1. The objective at this point is to climb, as expeditiously as possible, which is difcult because of the lack of excess thrust due to the large amount drag created by the gear and faps and the fact that only one engine is in operation. Therefore, the strategy is to retract the gear and faps as soon as possible. Since retracting faps at low speeds close to the ground is dangerous, the only option is to retract the gear.Once the gear is up and locked then the frst segment is fnished. During this segment the steady gradient of climb must be positive. SEGMENT 2The second segment starts at the end of the frst segment, i.e. when the gear is up. The objective now is to retract the faps. However, fap retraction is not permited below 400 ft, therefore the action by the pilot is simply to climb, at no less than V2, until 400 ft is reached. Once 400 ft is reached and fap retract can commence, segment 2 ends. Since the aeroplane has had the main source of drag removed, the minimum gradient requirement is more severe at no less than 2.4%.SEGMENT 3Segment three starts at or above 400 ft and is the fap retraction and acceleration segment. However, retracting the faps will increase the stall speed. This reduces the aeroplanes safety margin. Therefore, the aeroplane must accelerate during fap retraction from V2 to the zero fap speed and then to the fnal take-of speed. The fnal take-of speed is also called the fnal segment speed and is intended to be the one engine inoperative best angle of climb speed. Once this has happened, thrust can be reduced from maximum take-of thrust, TOGA, to maximum continuous thrust, MCT. In fact, maximum take-of thrust is limited to only 5 minutes, therefore, acceleration and fap retraction must be complete by then. The acceleration phase of this segment poses a climb problem. To climb, excess thrust is needed. However, accelerating the aeroplane also requires excess thrust. Therefore, it is not possible to adequately accelerate and climb at the same time and, as such, segment three can be fown level as shown by the illustration.SEGMENT 4The forth segment starts when the faps are retracted, the fnal segment speed is achieved and the thrust is set to maximum continuous thrust. From this point the aeroplane is climbed to above 1500 ft where the take-of fight path ends. The pilot will either rejoin the airfeld to land to continue on-route to the take-of alternate. The climb gradient for this last stage must not be less than 1.2%. One last point. The gradient requirements for Class A aeroplanes with more than two engines forms part of the syllabus and therefore they need to be remember as well. Although we have already mentioned it in the previous chapter, the Climb Limit Mass tells us what the maximum mass of the aeroplane can be to order achieve the minimum gradient requirements. Having a mass greater than the climb limit mass would mean that the aeroplane will not have sufcient performance in the event of engine failure to achieve the climb gradient requirements. The graph to calculate the climb limit mass is on page 11 of section 4 in CAP 698. We have worked through an example in the previous chapter, but none the less ensure youre confdent in being able to use this graph. The introduction to the climb section, shown on page 10 of section 4 reiterates the climb requirements and it states that during the take-of climb the aeroplane must:a. Atain the most severe gradient requirement of the take-of Net Flight Path and b. Avoid all obstacles in the obstacle accountability area by the statutory minimum vertical interval. 409Chapter 15Class ATake Off ClimbBoth these requirement have already been mentioned at the beginning of this chapter. The Climb Limit Mass deals with point a. In other words, being at or below the climb limit mass guarantees atainment of the most severe gradient requirement of the take-of fight path. What is the most severe angle of the take-of fight path. The most severe gradient requirement is in segment 2, which for a twin engine jet aeroplane is 2.4%. Therefore the climb limit mass ensures that the aeroplane, in the event of engine failure is able to atain a 2.4% gradient or more. Remember that these climb gradients are air based and therefore independent to the efect of wind. Looking through the climb limit mass graph confrms this since there is no account for wind.OBSTACLE CLEARANCEHaving discussed the minimum climb gradient requirements and how to ensure the aeroplane atains them, we need to focus on the obstacle clearance requirements. In JAR-OPS 1 it states that an operator must ensure that the net take-of fight path must clear all obstacles by a vertical margin of at least 35 ft. If the aeroplane is unable to do so it must turn away from the obstacle and clear it by a horizontal distance of at least 90 m plus 0.125 x D, where D is the horizontal distance the aeroplane has travelled from the end of the take-of distance available. 90m + 0.125 DFor aeroplanes with a wingspan of less than 60 m a horizontal obstacle clearance of half the aeroplane wingspan plus 60 m, plus 0.125 x D may be used. 60m + wing span + 0.125 DHowever, obstacles further away than the values shown below need not be considered.We have seen this horizontal clearance information before when we were discussing Class B multi engine obstacle clearance. However, there are two crucial points to consider when trying to work out the vertical clearance.The frst relates to climb gradient, which if you remember, is a ground based gradient. In order to work out the obstacle clearance of the aeroplane, the climb gradient needs to be known. But JAR-OPS 1 states that the climb gradient to use for the purpose of calculating obstacle clearance must be the net climb gradient. Remember that the net gradient is the gross gradient diminished by a safety factor. In this case the safety factor changes depending on the number of engines. The net gradient is the gross gradient reduced by:0.8% for 2 engined aircraft0.9% for 3 engined aircraft1.0% for 4 engined aircraft

410Chapter 15Class ATake Off ClimbUsing the net gradients instead of the gross gradient in the take-of fight path will produce a net take-of fight. As we stated at the beginning, the net take-of fight path must clear all obstacles by 35ft. The second point to consider is efect of wind on the ground gradient. Headwinds will increase the ground gradient and improve obstacle clearance. Where as tailwinds will decrease the ground gradient and deteriorate the obstacle clearance. JAR-OPS 1 states that when adjusting for wind to calculate the ground gradient, no more than 50% of the reported head wind and no less than 150% of the reported tailwind must be used. Carrying out this rule simply adds another safety margin into the calculation.We stated that if the aeroplane is unable to clear the obstacle vertically, then, it can turn away from the obstacle and clear it horizontally. However, there is restriction on how much the aeroplane is allowed to turn. Clearly it is not safe if the aeroplane needs to bank sharply to clear the obstacle by the regulatory margins. Turning can increase the efective weight by imposing extra g loads and therefore the climb gradient is reduced and stall speeds are increased. Allowance must be made for the efect of the turn on the climb gradient and speed. The fight manual usually gives a gradient decrement for a 15 banked turn at V2. For greater bank angles: For 20 bank, use 2 x gradient decrement and V2 + 5 kt For 25 bank, use 3 x gradient decrement and V2 + 10 ktTURNS ON THE FLIGHT PATH Turns are not allowed below a height of half the wingspan or 50ft whichever is greater. Up to 400ft, bank angle may not be more than 15. Above 400 ft, bank angle may not be more than 25.JAR-OPS 1 does permit operators to exceed these bank angles providing the operator uses special procedures and that these procedures have been approved by the relevant authority. The special procedures must take account of the gradient loss from such bank angles and these must be published in the aeroplane fight manual. The maximum bank angles that the special procedures allow is up to 20 between 200 and 400 ft and up to 30 between 400 and 1500 ft.If any turn of more than 15 is required at any point in the take-of fight path, then the vertical clearance is increased to 50 ft instead of 35 ft. Manually working out the obstacle clearance capability of the aeroplane could take a long time, since there are so many points to bear in mind and the calculation itself is quiet lengthy. Thankfully, most operators and manufactures have produced either rapid look up tables or graphs to quickly enable the pilot to work out if an obstacle in the take-of climb will be cleared by the relevant vertical margins following engine failure. These tables or graphs will produce a mass. This mass is called the obstacle limit mass and an example is shown in fgure 15.2 which can also be found in CAP 698 on page 36 and 37 of section 4. It is the maximum mass that will allow the aeroplane, in the event of engine failure to clear the obstacle by the relevant vertical margin.Notice that winds are included into the graph. This is important because obstacle clearance calculations must use ground gradients and these are dependent on wind. In fact, remember that JAR-OPS 1 had a rule about the wind. It stated that when using the winds to work out the ground gradient only use 50% of headwinds and no less than 150% tailwinds. Notice the slope of the headwind and tailwind lines. This shows that the graph applies the wind rule for you. Therefore, if you enter the graph with the actual reported wind the graph corrects it automatically so you dont need to.411Chapter 15Class ATake Off ClimbFigure 15.2 Obstacle Limit Mass412Chapter 15Class ATake Off ClimbOnce the details of the obstacle have been entered, as shown by the example, the obstacle limit mass can be read of, and, in the example on the graph itself the obstacle limit mass is 51,700 kg. Taking of with a higher mass than this would not ensure adequate vertical clearance. However, if this mass unduly restricts the take-of mass, then the aeroplane maybe dispatched with a greater mass so long as the aeroplane turns around the obstacle, clearing it by the relevant horizontal margins and does not exceed the turn restrictions when trying to do so.NOISE ABATEMENT PROCEDURESAeroplane operating procedures for the take-of climb shall ensure that the necessary safety of fight operations is maintained whilst minimizing exposure to noise on the ground.The frst procedure (NADP 1) is intended to provide noise reduction for noise-sensitive areas in close proximity to the departure end of the runway (see Figure 15.3). The second procedure (NADP 2) provides noise reduction to areas more distant from the runway end (see Figure 15.4). The two procedures difer in that the acceleration segment for fap/slat retraction is either initiated prior to reaching the maximum prescribed height or at the maximum prescribed height. To ensure optimum acceleration performance, thrust reduction may be initiated at an intermediate fap seting.413Chapter 15Class ATake Off ClimbNADP 1 This procedure involves a power reduction at or above the prescribed minimum altitude and the delay of fap/slat retraction until the prescribed maximum altitude is atained. At the prescribed maximum altitude, accelerate and retract faps/slats on schedule whilst maintaining a positive rate of climb, and complete the transition to normal en-route climb speed.The noise abatement procedure is not to be initiated at less than 800 ft above aerodrome elevation. The initial climbing speed to the noise abatement initiation point shall not be less than V2 plus 20 km/h (10 kt). On reaching an altitude at or above 800 ft above aerodrome elevation, adjust and maintain engine power/thrust in accordance with the noise abatement power/thrust schedule provided in the aircraft operating manual. Maintain a climb speed of V2 plus 20 to 40 km/h (10 to 20 kt) with faps and slats in the take-of confguration.At no more than an altitude equivalent to 3,000 ft above aerodrome elevation, whilst maintaining a positive rate of climb, accelerate and retract faps/slats on schedule. At 3,000 ft above aerodrome elevation, accelerate to en-route climb speed.Figure 15.4 Noise Abatement Take-Off Climb - Example of a Procedure AlleviatingNoise Close to the Aerodrome (NADP 1)Figure 15.3 Noise Abatement Take-Of Climb - Example of a Procedure AlleviatingNoise Close to the Aerodrome (NADP 1)414Chapter 15Class ATake Off ClimbNADP 2This procedure involves initiation of fap/slat retraction on reaching the minimum prescribed altitude. The faps/slats are to be retracted on schedule whilst maintaining a positive rate of climb. The power reduction is to be performed with the initiation of the frst fap/slat retraction or when the zero fap/slat confguration is atained. At the prescribed altitude, complete the transition to normal en-route climb procedures.The noise abatement procedure is not to be initiated at less than 800 ft above aerodrome elevation. The noise abatement procedure is not to be initiated at less than 800 ft above aerodrome elevation. The initial climbing speed to the noise abatement initiation point is V2 plus 20 to 40 km/h (10 to 20 kt). On reaching an altitude equivalent to at least 800 ft above aerodrome elevation, decrease aircraft body angle/angle of pitch whilst maintaining a positive rate of climb, accelerate towards VZF and either: reduce power with the initiation of the frst fap/slat retraction; or reduce power after fap/slat retraction.Maintain a positive rate of climb, and accelerate to and maintain a climb speed of VZF + 20 to 40 Km/h (10 to 20 kt) to 3,000 ft above aerodrome elevation. On reaching 3,000 ft above aerodrome elevation, transition to normal en-route climb speed.Figure 15.5 Noise Abatement Take-Off Climb - Example of a Procedure AlleviatingNoise Distant from the Aerodrome (NADP 2)Figure 15.4 Noise Abatement Take-Of Climb - Example of a Procedure AlleviatingNoise Distant from the Aerodrome (NADP 2)415Chapter 15Class ATake Off ClimbQUESTIONS1. The net fight path climb gradient after take-of compared to the gross climb gradient is:a. larger.b. equal.c. depends on type of aircraft.d. smaller.2. Given that the characteristics of a three engine turbojet aeroplane are as follows:Thrust = 50,000 N per engineg = 10 m/sDrag = 72,569 NMinimum gross gradient (2nd segment) = 2.7%The maximum take-of mass under segment two conditions in the net take-of fight path conditions is:a. 101,596 kgb. 286,781 kgc. 74,064 kgd. 209,064 kg3. During the fight preparation the climb limited take-of mass (TOM) is found to be much greater than the feld length limited TOM using 5 fap. In what way can the performance limited TOM be increased? There are no limiting obstacles.a. By selecting a higher fap seting.b. By selecting a higher V2.c. By selecting a lower V2.d. By selecting a lower fap seting.4. An operator shall ensure that the net take-of fight path clears all obstacles. The half-width of the obstacle-corridor at the distance D from the end of the TODA is at least:a. -90m + 1.125Db. 90m + D/0.125c. 90m + 0.125Dd. 0.125D5. The climb gradient is defned as the ratio of:a. true airspeed to rate of climb.b. rate of climb to true airspeed.c. the increase of altitude to horizontal air distance expressed as a percentage.d. the horizontal air distance over the increase of altitude expressed as a percentage.6. Which of the following statements, concerning the obstacle limited take-of mass for performance class A aeroplane, is correct?a. It should not be corrected for 30 bank turns in the take-of path.b. It should be calculated in such a way that there is a margin of 50 ft with respect to the net take of fight path.c. It cannot be lower than the corresponding climb limited takeof mass.d. It should be determined on the basis of a 35 ft obstacle clearance with the respect to the net take-of fight path.416Chapter 15Class ATake Off Climb7. The determination of the maximum mass on brake release, of a certifed turbojet aeroplane with 5, 15 and 25 faps angles on take-of, leads to the following values, with wind:Flap angle: 5 15 25Runway limitation: 66 000 69 500 71 500 2nd segment slope limitation: 72 200 69 000 61 800Wind correction: Head wind: + 120 kg per kt OR Tail wind: - 360 kg per kt Given that the tail wind component is equal to 5 kt, the maximum mass on brake release and corresponding fap angle will be:a. 67 700 kg / 15 degb. 69 000 kg / 15 degc. 72 200 kg / 5 degd. 69 700 kg / 25 deg8. The requirements with regard to take-of fight path and the climb segments are only specifed for:a. the failure of two engines on a multi-engined aeroplane.b. the failure of the critical engine on a multi-engines aeroplane.c. the failure of any engine on a multi-engined aeroplane.d. 2 engined aeroplane.9. The frst segment of the take-of fight path ends:a. at completion of gear retraction.b. at completion of fap retraction.c. at reaching V2.d. at 35 ft above the runway.10. Which statement, in relation to the climb limited take-of mass of a jet aeroplane, is correct?a. 50% of a head wind is taken into account when determining the climb limited take-of mass.b. On high elevation airports equipped with long runways the aeroplane will always be climb limited.c. The climb limited take-of mass decreases with increasing OAT.d. The climb limited take-of mass is determined at the speed for best rate of climb.11. Which one of the following is not afected by a tail wind?a. the feld limited take-of mass.b. the obstacle limited take-of mass.c. the take-of run.d. the climb limited take-of mass.12. How does TAS vary in a constant Mach climb in the troposphere?a. TAS increases.b. TAS is constant.c. TAS is not related to Mach Number.d. TAS decreases.417Chapter 15Class ATake Off Climb13. For this question use Figure 4.5 in CAP 698 Section 4.With regard to the take-of performance of a twin jet aeroplane, why does the take-ofperformance climb limit graph show a kink at 30C, pressure altitude 0 ft?a. At lower temperatures one has to take the danger of icing into account.b. The engines are pressure limited at lower temperatures, at higher temperatures they are temperature limited.c. At higher temperatures the fat rated engines determines the climb limit mass.d. At higher temperatures the VMBE determines the climb limit mass.14. Which of the following sets of factors will increase the climb-limited take-of mass (TOM)?a. High fap seting, low PA, low OAT.b. Low fap seting, high PA, high OAT.c. Low fap seting, high PA, low OAT.d. Low fap seting, low PA, low OAT.15. For this question use Figure 4.5 in CAP698 Section 4. Consider the take-of performance for the twin jet aeroplane climb limit chart. Why has the wind been omited from the chart?a. There is a built-in safety measure.b. The climb limit gradient requirements are taken relative to the air.c. The efect of the wind must be taken from another chart.d. There is no efect of the wind on the climb angle relative to the ground.16. Which of the following statements is applicable to the acceleration height at the beginning of the 3rd climb segment?a. The minimum legally allowed acceleration height is at 1500 ft.b. There is no minimum climb performance requirement when fying at the acceleration height.c. The minimum one engine out acceleration height must be maintained in case of all engines operating.d. The maximum acceleration height depends on the maximum time take-of thrust may be applied.17. In relation to the net take-of fight path, the required 35 ft vertical distance to clear all obstacles is:a. based on pressure altitudes.b. the height by which acceleration and fap retraction should be completed.c. the height at which power is reduced to maximum climb thrust.d. the minimum vertical distance between the lowest part of the aeroplane and all obstacles within the obstacle corridor.18. Which of the following statements is correct?a. The performance limited take-of mass is independent of the wind component.b. The accelerate stop distance required is independent of the runway condition.c. The take-of distance with one engine out is independent of the wind component.d. The climb limited take-of mass is independent of the wind component.418Chapter 15Class ATake Off Climb19. Which of the following statements with regard to the actual acceleration height at the beginning of the 3rd climb segment is correct?a. The minimum value according to regulations is 1000 ft.b. There is no legal minimum value, because this will be determined from case to case during the calculation of the net fight path.c. The minimum value according to regulations is 400 ft.d. A lower height than 400 ft is allowed in special circumstances e.g. noise abatement.20. For take-of obstacle clearance calculations, obstacles in the frst segment may be avoided:a. by banking not more than 15 between 50 ft and 400 ft above the runway elevation.b. by banking as much as needed if aeroplane is more than 50 ft above runway elevation.c. only by using standard turns.d. by standard turns but only after passing 1,500 ft.21. The second segment begins:a. when landing gear is fully retracted.b. when fap retraction begins.c. when faps are selected up.d. when acceleration starts from V2 to the speed for fap retraction.22. At which minimum height will the second climb segment end?a. 1,500 ft above feld elevation.b. 400 ft above feld elevation.c. 35 ft above ground.d. When gear retraction is completed.23. On a segment of the take-of fight path an obstacle requires a minimum gradient of climb of 2.6% in order to provide an adequate margin of safe clearance. At a mass of 110,000 kg the gradient of climb is 2.8%. For the same power and assuming that the angle of climb varies inversely with mass, at what maximum mass will the aeroplane be able to achieve the minimum gradient?a. 121,310 kgb. 106,425 kgc. 118,455 kgd. 102,142 kg24. During take-of, the third segment begins:a. when acceleration to fap retraction speed is started.b. when landing gear is fully retracted.c. when acceleration starts from VLOF to V2.d. when fap retraction is completed.25. If there is a tail wind, the climb limited take-of mass will:a. increase.b. decrease.c. increase in the faps extended case.d. not be afected.419Chapter 15Class ATake Off Climb26. A higher pressure altitude at ISA temperature:a. has no infuence on the allowed take-of mass.b. decreases the feld length limited take-of mass.c. decreases the take-of distance.d. increases the climb limited take-of mass.27. The minimum climb gradient required on the 2nd fight path segment after the take-of of a jet aeroplane is defned by the following parameters:1 Gear up 2 Gear down 3 Wing faps retracted 4 Wing faps in take-of position 5 All Engines at the take-of thrust 6 One engine inoperative, remainder at the take-of thrust 7 Speed equal to V2 + 10 kt 8 Speed equal to 1.3 VS 9 Speed equal to V2 10 At a height of 35 ft above the runwayThe correct statements are:a. 2, 3, 6, 9 b. 1, 4, 5, 10 c. 1, 5, 8, 10 d. 1, 4, 6, 928. In the event that the take-of mass is obstacle limited and the take-of fight path includes a turn, the maximum bank angle is:a. 10 degrees up to a height of 400 ft.b. 20 degrees up to a height of 400 ft.c. 25 degrees up to a height of 400 ft.d. 5 degrees up to height of 400 ft.29. Up to which height in noise abatement departure procedure 1 (NADP1) must V2 + 10 to 20 kt be maintained?a. 1,500ft.b. 3,000ft.c. 800ft.d. 500ft.30. Reference point zero refers to the:a. Point where the aircraft lifts of the ground.b. Point where the aircraft reaches V2.c. Point on the ground where the aircraft reaches 35ft.d. Point where gear is selected up.31. A Boeing 737 has a wingspan of 28.9 m. An obstacle is situated at a distance of 4,264 ft from the end of the TODA. What is the minimum horizontal obstacle clearance?a. 607.45 mb. 252.50 mc. 236.95 md. 240 m420Chapter 15Class ATake Off ClimbANSWERS1. D 11. D 21. A 31. C2. A 12. D 22. B3. A 13. C 23. D4. C 14. D 24. A5. C 15. B 25. D6. D 16. B 26. B7. A 17. D 27. D8. B 18. D 28. D9. A 19. C 29. B10. C 20. A 30. C421Chapter 16Class AEn RouteCHAPTER SIXTEENCLASS A: EN ROUTEContentsEN-ROUTE PHASE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423CLIMB PROFILE / CLIMB SCEDULE . . . . . . . . . . . . . . . . . . . . . . . . . . . 423CRUISE SPEEDS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425COST INDEX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 426CRUISE ALTITUDES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 428AERODYNAMIC CEILING AND MANOEUVRE CEILING . . . . . . . . . . . . . . 429NORMAL DESCENT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 430EMERGENCY DESCENT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431OBSTACLE CLEARANCE REQUIREMENTS . . . . . . . . . . . . . . . . . . . . . . 431RANGE LIMIT FOLLOWING ENGINE FAILURE. . . . . . . . . . . . . . . . . . . . 434ETOPS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435QUESTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437ANSWERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444422Chapter 16Class AEn Route423Chapter 16Class AEn RouteEN-ROUTE PHASEThe en-route phase of fight starts at 1500 ft above the departure aerodrome and ends once the aeroplane has reached 1500 ft above intended destination aerodrome. Similarly with other phases of fight for Class A aeroplanes, the en-route regulations account for engine failure. Therefore, manufacturers and operators of Class A aeroplanes must ensure that the performance of the aeroplane subsequent to engine failure is still able to meet the regulations. Firstly we will discus the climb to the en-route altitude, then we will detail the en-route altitudes and how they are calculated as well as discussing various fight speeds. Lastly we will detail the descent, both normal descent, and the descent forced by either engine failure or depressurization.CLIMB PROFILE / CLIMB SCHEDULEAfter a normal take-of, climbing to the en-route altitude is a straight forward afair. Once the aeroplane confguration is clean, a set climb profle or climb schedule will be fown. Initially the aeroplane climbs at a constant indicated airspeed. However, continuously climbing at a constant indicated airspeed causes the Mach number to rise. Beyond a certain altitude, the Mach number gets too high and serious aerodynamic forces start to afect the aeroplane. In the 737 family the maximum Mach number, MMO is 0.82. Therefore, at some lower altitude the aeroplane needs to change its climb profle to a constant Mach number climb. The altitude at which this change occurs is called the cross over or change-over altitude.In brief summary then, the climb profle involves the aeroplane initially climbing at a constant indicated airspeed, and then at the cross over altitude, the aeroplane then climbs at a constant mach number. However, ICAO limits the maximum indicated airspeed to 250 knots below 10,000 ft. For the majority of the 737 family, the climb profle is 250 knots indicated airspeed up to 10,000 ft, then the aeroplane is accelerated to 280 knots and the climb continued at 280 knots. As the aeroplane climbs, Mach number will increase and when the Mach number reaches 0.74, the aeroplane maintains a climb speed of 0.74 until the en-route cruise altitude. Using Figure 16.1 you will notice that the cross over altitude is at about 25,700 ft.424Chapter 16Class AEn RouteFigure 16.1 A typical climb schedule for a 737 400.However, you need to understand what happens to the crossover altitude, if a faster indicated airspeed climb is maintained, and for this example well increase the second part of the indicated airspeed climb from 280 to 325 knots. Looking at Figure 16.2, initially the aeroplane will still climb at 250 knots, but, once reaching 10,000 ft the aeroplane will be accelerated to 325 knots and then climb at 325 knots. Once the Mach number has increased to 0.74, then the aeroplane climbs at 0.74. Notice now the crossover altitude is lower at 18,000 ft. Therefore if the indicated airspeed is increased in the climb profle, the crossover altitude decreases to a lower altitude.425Chapter 16Class AEn RouteFigure 16.2 A typical climb schedule using a faster indicated airspeed.CRUISE SPEEDSBefore we proceed to discuss the en-route altitude, it is important at this point to briefy detail a few important speeds, most of which you have already covered. These speeds are commonly expressed as Mach numbers. The frst of these speeds is the maximum operating speed, either called VMO when using indicated airspeed or MMO when using mach numbers. For the majority of the 737 family VMO is 340 knots and MMO is 0.82. Flying beyond these speeds in a commercial operational context is not permited and may cause either structural damage or a high speed stall.The next two speeds are used in reference to describing the range of the aeroplane and were mentioned in the en-route chapter of the general performance principles section. These two speeds are the maximum range cruise speed, MRC, and long range cruise speed, LRC. When these speeds are referenced in terms of a mach number the abbreviation is changed to MMR and MLRC respectively. 426Chapter 16Class AEn RouteFigure 16.3 A graph showing the relation between the long range cruise speed (LRC) and the maximum range cruise speed (MRC)As you can see from Figure 16.3, ploting cost and speed, the advantage of fying at the maximum range speed is simply that the aeroplane will use the least amount of fuel and therefore have the least fuel cost for a given distance. However, operationally, the faster long range cruise is used. The simple reason why this speed is used is because by geting to destination more quickly; more revenue earning fights can be carried out in any give period. In other words over a given time period, 4% more fights can be carried out with only a fuel consumption increase of 1%. The long range cruise speed does sufer from limitations. It doesnt take into account the variable cost of fuel from day to day or month to month and neither does it account for the operational costs. When fuel prices are high, the extra fuel consumption may dramatically increase the overall cost of the fight and a more operationally economical speed may need to be fown. The relationship of these costs is explained by the use of a cost index and the speed fown based upon the cost index this is called ECON. This will be discussed next.COST INDEXThe fundamental rationale of the cost index concept is to achieve minimum operation trip cost by means of a trade-of between time related costs and fuel related costs. The cost index is used to take into account the relationship between fuel-and time-related costs. With time-related costs, the faster the aircraft is fown, the more money is saved in time costs. This is because the faster the aircraft is fown, the more miles can be fown for time-related components. It also means that more miles can be fown between inspections when considering maintenance costs. These costs are minimum at the maximum operating speed VMO/MMO. However, if the aircraft is fown at such a high speed the fuel burn increases and total fuel cost for the trip increases. Fuel costs on the other hand will be minimum at the maximum range cruise speed (MRC) and maximum and the maximum operating speed. Adding the time related costs and fuel related costs together, produces a direct operating cost, or more simplistically, a total operating cost. The fight management system uses the time and fuel related costs help select the best speed to fy. 427Chapter 16Class AEn RouteFigure 16.4 A graph ploting the cost of the aeroplane against the aeroplane speedLooking at Figure 16.4 you can see from the total cost curve that the speed which gives the minimum total operating cost is the most economical speed to fy. This speed is called ECON, in other words the minimum cost speed. The value of the ECON speed is worked out by the fight management system based upon the value of the cost index. As a formula the cost index is a ratio of cost of time, CT, divided by the cost of fuel, CF.When fuel costs are high and time cost are very low, the cost index would be almost zero and the blue total cost line is moved to the left. The intersection point of the other cost lines will lie very close to the maximum range cruise speed and. With a cost index of zero, the ECON speed (found at the botom of the blue line) would now be at the maximum range speed. When time costs are high and fuel costs are low, the cost index would be very high and the blue total cost line moves to the far right of the graph. The ECON speed, found at the botom of the blue curve would now be very close to the maximum operating speed.To summarise then, increasing the cost index from zero to maximum will increase the ECON speed from the maximum range speed to maximum operating speed. For most aeroplanes the cost index varies from zero to 99 or from zero to 999.428Chapter 16Class AEn RouteCRUISE ALTITUDESOnce the aeroplane has completed the climb profle and has reached the top of the climb, the aeroplane will level of at the appropriate altitude. This cruise altitude should ideally coincide with optimum altitude. You may recall that the optimum altitude was the altitude for maximum specifc range or maximum fuel mileage. As a general rule, this altitude is not constant. As weight decreases during the fight from fuel consumption, the optimum altitude increases, as illustrated in the Figure 16.5. Figure 16.5 A graph showing important altitudes for a typical medium range jetTherefore, to constantly fy at the optimum altitude, the aeroplane will not actually fy level, but will in fact slowly be climbing in the cruise. On the other hand, ATC restrictions require level fight cruise to ensure vertical separations with other aeroplanes. To try an accommodate ATC in congested airspace aircraft must fy by segments of constant altitude which must be as close as possible to the optimum altitude. The level segments are established within 2,000 feet from the optimum altitude. The procedure is called the step climb which you have seen in general principles chapter on En-route. Remaining within 2000 ft of the optimum altitude ensures the range is 99% of the maximum specifc range. There may be several step climbs during the fight and the aeroplane will be gaining altitude throughout this process. However, there is a limit to how high the aeroplane is permited to operate and able to operate.429Chapter 16Class AEn RouteAs the aeroplane altitude increases, the thrust that is required to maintain a given speed, increases. Eventually, there will be an altitude where the thrust is increased to its maximum cruise value and it would not be possible to climb any higher without exceeding the thrust limits. This altitude is called the Maximum Altitude and is shown in Figure 16.5. However, the hoter the atmosphere, the lower this altitude becomes and in exceptionally hot atmospheres the maximum altitude is almost the same as the optimum altitude. Although the aeroplane cannot operate above the maximum altitude, there are other altitude limits placed upon the aeroplane.AERODYNAMIC CEILING AND MANOEUVRE CEILINGBefore we can discuss the other limits on the operating altitude of an aeroplane it is important to discus aeroplane stalling. When the speed of the aeroplane is reduced, in order to still produce enough lift to balance weight, the angle of atack must increase. However, below a certain speed, the angle of atack on the wings is such that the fow of air over the wing starts to separate from the boundary layer producing turbulent air fow. The separation point fuctuates back and forth along the wing making strong eddies in the turbulent airfow. These strong eddies, bufet the elevators or tail plane. This phenomenon is called the low speed bufet. Flying below this speed will dramatically decrease the lift and a full stall ensues.

Figure 16.6 The relationship of the mach number for the low speed and high speed bufets with altitudeFor a given weight and confguration, the aeroplane will always stall at the same indicated airspeed but the equivalent mach number for the low speed bufet and stall increases with altitude as illustrated by a simplistic graph shown in Figure 16.6. The Mach number for the low speed bufet is abbreviated to MMIN. A similar bufet can occur at high speed. At very high speeds, close to the speed of sound, the compressibility of the air ahead of the aeroplane leads to the formation of shock waves or high pressure waves. These shock waves create a disturbance to the fow of air over the wing causing it to separate and create turbulent eddies. Similarly to the low speed bufet these eddies will bufet the elevator. This phenomenon is called the high speed bufet and the Mach number for the high speed bufet decreases with altitude. Flying faster than this speed may cause a high speed shock stall in an aeroplane whose wings are not designed to overcome such efects. The Mach number for the high speed bufet decreases with 430Chapter 16Class AEn Routealtitude, as shown. This speed is commonly abbreviated to MMAX and is shown by the backward sloping red line to the right of the graph in Figure 16.6.Taking into consideration both the Mach number for low speed and high speed bufet, it means that there are two mach numbers, below and above which the aeroplane is unable to fy. This speed range between the Mach number for the low speed stall and high speed stall is called the bufet margin. The bufet margin is the speed range between the low and speed and high speed bufet. The important point to understand is that the margin between the low speed and high speed bufet decreases with altitude. Notice that there is an altitude with that the low speed and high speed bufets are coincident at the same velocity. It is impossible to fy higher than this altitude. Flying slower or faster than the speed shown will stall the aeroplane. In fact, even manoeuvring the aeroplane will initiate a stall because manoeuvring the aeroplane will increase the efective weight and increase the stall speed. This altitude is called the aerodynamic ceiling, or cofn corner. To prevent aeroplanes from operating too close to this altitude, an operational limit is set below this point. Aerodynamic ceiling is the altitude where the low speed and high speed bufets are coincident. Notice that a 1.3 g manoeuvre moves the bufet speed lines to the faded red position in Figure 16.6. Notice that now, the Mach number for the low and speed and high speed bufet are coincident at a lower altitude. This altitude is called the 1.3 g bufet limit altitude or manoeuvre ceiling and is usually about 4,000 to 6,000 ft below the aerodynamic ceiling.NORMAL DESCENTWhen the aeroplane gets close to the destination airfeld it will reach a point which marks the beginning of the descent. This is called the top of descent. You may remember that in order to initiate a descent, frstly the thrust must be reduced, and then the nose is lowered to get weight to act forwards to balance the drag. The balance of forces ensures a constant speed can be maintained during the descent. The descent profle is almost the reverse of the climb profle. The climb for a typical 737 is initially fown at 250 knots, then at 10,000 ft this changes to 280 knots and then at the cross over altitude mach 0.74 is maintained. The descent is fown initially and 0.74 mach, then at the cross over altitude the speed in kept constant at 280 knots, but when 10,000ft is reached no more than 250 knots must be fown.Figure 16.7 A typical descent profle for a medium range jet431Chapter 16Class AEn RouteShown in Figure 16.7 are the characteristics of the descent so you can see what happens to the gradient and rate of descent throughout the descent profle. If at any point air trafc control asks the pilots to expedite the descent, then the only action by the pilots would be to deploy the speeds brakes. This increases the drag, which must be balanced by more weight apparent thrust, therefore the nose is lowered which increases both the angle and rate of descent as per the instruction of air trafc control. EMERGENCY DESCENTThe next descent to consider is the descent characteristic following either depressurisation or engine failure. In fight, engine or pressurization failures force a premature descent and therefore the performance becomes very constraining over mountainous areas. In case of an engine failure during fight, the remaining thrust is no longer sufcient to balance the drag force and therefore the cruise speed cannot be maintained. The only solution is to descend to a lower fight altitude, where the remaining engine can provide enough thrust to balance the drag and allow level fight once more. To achieve this, the aeroplane is allowed to decelerate from the selected cruise speed to the velocity of minimum drag, and then the nose is lowered to get weight apparent thrust to balance drag. As the aeroplane descends into the lower atmosphere where density is greater, the remaining engine can develop more thrust which will mean that weight apparent thrust can be slowly reduced until an altitude is reached where the remaining engine generates enough thrust to balance drag. At this altitude the aeroplane can level of.

This procedure is called the drift down, and it produces a Driftdown profle. This path must of course, be above all relevant obstacles, but this will be discussed later.With a pressurization failure the procedure is litle diferent. At high altitude, following depressurization, the air in the cabin escapes and very quickly the cabin air becomes the air same as outside. At high altitude the problem is that there is very litle oxygen to breathe. Therefore oxygen must carried on board and supplied to crew and passengers through oxygen masks. However, the amount of oxygen carried is limited, therefore the aeroplane must descend, as rapidly as possible to 10,000 ft where there is sufcient oxygen, before the oxygen supply runs out. The procedure involves confguring the aeroplane for the maximum rate of descent. You may recall that in order to achieve a maximum rate of descent, excess power required has to be a large as possible. Therefore drag must be high and speed must be high. As a result, the frst actions of the pilot are to don the oxygen masks, initiate descent, close the throtles apply the speed brakes and then allow the aeroplane to accelerate to maximum operating speed which is either VMO or MMO. This confguration is then maintained till at least 10,000 ft, or minimum safe en-route altitude, where there is sufcient oxygen to breathe.OBSTACLE CLEARANCE REQUIREMENTSOne of crucial points about the driftdown procedure is the clearance of obstacles. Because the aeroplane is forced to descend, terrain, like mountains may present a fight hazard. When assessing the terrain hazard a safety margin must be introduced. When planning routes and planning the fight profle, it is not the gross fight profle that must assumed, but rather the net fight profle. In other words, the fight profle must be made worse by a safety factor. This safety factor is based on assuming a gradient of descent that is worse than the aeroplane can actually achieve. For two engine aeroplanes with one engine inoperative the gross gradient of descent is increased by 1.1%. This increases to 1.4% for 3 engine aeroplanes and 1.6% for four engine aeroplanes.432Chapter 16Class AEn RouteFigure 16.8 Net and Gross Descent Profles for a typical twin engine medium range jetJAR OPS 1 details the regulations for en-route obstacle clearance. It states that the gradient of the net fight path must be positive at at least 1000 ft above all terrain and obstructions along the route within 5 nm on either side of the intended track. From Figure 16.8 you can see that in the example the net fight path does indeed become positive 1000 ft above the obstacle. This regulation is essentially saying that the aeroplane must be able to demonstrate it can level of and achieve a positive climb at 1000 ft above all obstacles along the route. This maybe be very difcult to achieve if there was a very high mountain around.If the aeroplane is unable to have a positive climb 1000 ft above all obstacles, then the aeroplane may level of at a lower altitude so long as any obstacle encountered in the descend is cleared vertically by at least 2000 ft. Using Figure 16.8, if the net fight path cannot be positive 1000 ft above the obstacle then the aeroplane may descend lower, BUT, the fight path must not enter the 2000 ft zone above the terrain.433Chapter 16Class AEn RouteIn order to fnd out if the aeroplane is able to level of at 1000ft above an obstacle, then use the graph in CAP 698 on page 40 of section 4. This has been reproduced in Figure 16.9. Figure 16.9 A graph to calculate the maximum mass for a given net level of altitudeSimply enter the graph with level of altitude that is required (obstacle height AMSL + 1000 ft) and the graph will show the mass that the aeroplane will need to be at in order to level of at that altitude. In the example on the graph, in order to level of at 18,000 ft in an ISA + 20 atmosphere the aeroplane would need to have a mass of just less than 48,000 kg. If this is not possible, then other graphs, shown on page 41 to 44 of CAP698 must be used. 434Chapter 16Class AEn RouteThese graphs are more complicated, but importantly, these graphs will enable the pilot to work out if the aeroplane can clear any obstacle in the fight path by 2000 ft. When you use the graph be careful to adjust the weight of the aeroplane for any non standard conditions and anti ice use. Notice that the heavier the aeroplane the longer and lower the driftdown procedure is. The last of the regulations state that the aeroplane must have a positive gradient at 1,500 ft above the aerodrome where the landing is assumed to be made after engine failure.RANGE LIMIT FOLLOWING ENGINE FAILUREAfter engine failure, the lower operating altitude signifcantly decreases the engines efciency. So much so that the fuel-fow on the remaining engine is almost as much as the fuel fow with both engines operating at high altitude. This fact, together with the reduced true airspeed, means that the specifc range is dramatically decreased. As a result of the reduced range it is now not possible to reach the destination airfeld and in fact the priority now is to fnd an alternate airfeld to land before the fuel runs out. This issue is of such importance that is was necessary to regulate it. The authorities have to set a safety standard that in the event of engine failure the aeroplane must have the capability of reaching a suitable airfeld within a certain time period.JAR-OPS 1 states that twin engine aeroplanes beyond a certain size must be no further away from a suitable aerodrome than the distance fown in 60 minutes using the one engine operative cruise speed as TAS in still air. For aeroplanes with 3 or more engines the time is increased to 90 minutes. Therefore at all points on the route, a twin engine aeroplane must be within 60 minutes of an alternate airfeld. This regulation has a signifcant impact on fight routes, especially over the sea. In the example in Figure 16.10 you can see that to comply with the 60 minute the rule the aeroplane track must all times be within the 60 minute range limit of a suitable alternate airfeld. From the diagram in Figure 16.10, a direct track to North America from Europe is not possible.Figure 16.10 Range Limits on twin engine medium range jets.435Chapter 16Class AEn RouteHowever, as more and more reliable and efcient aeroplanes are produced, an extension to the 60 minute rule has been introduced.ETOPSThe extension to the 60 minute rule is called ETOPS and it stands for Extended range with Twin engine aircraft OPerationS. It hugely increases the operational capability of twin engine aeroplanes where only 3 or more engine aeroplanes could operate. However, ETOPS must be applied for by the airlines concerned and approval gained from the appropriate aviation authority. To gain ETOPS approval, a greater range of performance parameters must be known and these accompany the application and are eventually published in the operating manual. These include extra data for Area of operation Critical Fuel Reserves Net Level-of AltitudesGaining an ETOPS approval of 120 minutes for example will greatly beneft fight tracks across the Atlantic Ocean, as shown in Figure 16.10. The route from Paris to New York for example, can now be fown direct by a twin jet aeroplane. Currently, the longest ETOPS approval is given to the Boeing 777. It holds an approval of 180 minutes with contingencies for 207 minutes over the Pacifc.In the future ETOPS may be evolving into a newer system, called LROPS. LROPS stands for Long Range Operational Performance Standards, which will afect all aircraft, not just those with a twin-engine confguration.436Chapter 16Class AEn Route437Chapter 16Class AEn RouteQUESTIONS1. Which statement with respect to the step climb is correct?a. A step climb provides beter economy than a cruise climb.b. Performing a step climb based on economy can be limited by the 1.3g altitude.c. In principle a step climb is performed immediately after the aircraft has exceeded the optimum altitude.d. A step climb may not be performed unless it is indicated in the fled fight plan.2. Which statement with respect to the step climb is correct?a. A step climb is executed because ATC desires a higher altitude.b. A step climb is executed in principle when, just after levelling of, the 1.3g altitude is reached.c. Executing a desired step climb at high altitude can be limited by bufet onset at g-loads larger than 1.d. A step climb must be executed immediately after the aeroplane has exceeded the optimum altitude.3. The lift coefcient decreases during a glide at a constant Mach number, mainly because the:a. TAS decreases.b. glide angle increases.c. IAS increases.d. aircraft mass decreases.4. An aeroplane carries out a descent from FL 410 to FL 270 at cruise Mach number, and from FL 270 to FL 100 at the IAS reached at FL 270. How does the angle of descent change in the frst and in the second part of the descent? Assume idle thrust and clean confguration and ignore compressibility efects. a. Increases in the frst part; is constant in the second.b. Increases in the frst part; decreases in the second.c. Is constant in the frst part; decreases in the second.d. Decreases in the frst part; increases in the second.5. During a glide at constant Mach number, the pitch angle of the aeroplane will:a. decrease.b. increase.c. increase at frst and decrease later on.d. remain constant.6. If a fight is performed with a higher Cost Index at a given mass which of the following will occur?a. A beter long range.b. A higher cruise mach number.c. A lower cruise mach number.d. A beter maximum range.438Chapter 16Class AEn Route7. During a descent at constant Mach Number, the margin to low speed bufet will:a. decrease, because the lift coefcient decreases.b. increase, because the lift coefcient decreases.c. remain constant, because the Mach number remains constant.d. increase, because the lift coefcient increases.8. A jet aeroplane is climbing with constant IAS. Which operational speed limit is most likely to be reached?a. The Stalling speed.b. The Minimum control speed air.c. The Mach limit for the Mach trim system.d. The Maximum operating Mach number.9. ETOPS fight is a twin engine jet aeroplane fight conducted over a route, where no suitable airport is within an area of:a. 75 minutes fying time at the approved one engine out cruise speed.b. 60 minutes fying time in still air at the approved one engine out cruise speed.c. 60 minutes fying time in still air at the normal cruising speed.d. 30 minutes fying time at the normal cruising speed.10. The danger associated with low speed and/or high speed bufet:a. limits the manoeuvring load factor at high altitudes.b. can be reduced by increasing the load factor.c. exists only above MMO.d. has to be considered at take-of and landing.11. Which data can be extracted from the Bufet Onset Boundary Chart?a. The value of the Mach number at which low speed and shock stall occur at various weights and altitudes.b. The values of the Mach number at which low speed bufet and Mach bufet occur at various masses and altitudes.c. The value of maximum operating Mach number (MMO) at various masses and power setings.d. The value of the critical Mach number at various masses and altitudes.12. Which of the following factors determines the maximum fight altitude in the Bufet Onset Boundary graph?a. Aerodynamics.b. Theoretical ceiling.c. Service ceiling.d. Economy.13. The optimum cruise altitude increases:a. if the aeroplane mass is decreased.b. if the temperature (OAT) is increased.c. if the tailwind component is decreased.d. if the aeroplane mass is increased.439Chapter 16Class AEn Route14. Which of the following statements with regard to the optimum cruise altitude (best fuel mileage or range) is correct?a. An aeroplane usually fies above the optimum cruise altitude, as this provides the largest specifc range.b. An aeroplane sometimes fies above or below the optimum cruise altitude, because ATC normally does not allow aeroplanes to fy continuously at the optimum cruise altitude.c. An aeroplane always fies below the optimum cruise altitude, as otherwise Mach bufet can occur.d. An aeroplane always fies on the optimum cruise altitude, because this is most atractive from an economy point of view.15. The optimum altitude:a. is the altitude up to which cabin pressure of 8 000 ft can be maintained.b. increases as mass decreases and is the altitude at which the specifc range reaches its maximum.c. decreases as mass decreases.d is the altitude at which the specifc range reaches its minimum.16. Under which condition should you fy considerably lower (4,000 ft or more) than the optimum altitude?a. If, at the lower altitude, either more headwind or less tailwind can be expected.b. If, at the lower altitude, either considerably less headwind or considerably more tailwind can be expected.c. If the maximum altitude is below the optimum altitude.d. If the temperature is lower at the low altitude (high altitude inversion).17. What happens to the specifc range with one or two engines inoperative?a. At frst improved and later reduced.b. It decreases.c. It increases.d. Unafected by engine failure.18. The optimum cruise altitude is:a. the pressure altitude up to which a cabin altitude of 8000 ft can be maintained.b. the pressure altitude at which the speed for high speed bufet as TAS is a maximum.c. the pressure altitude at which the highest specifc range can be achieved.d. the pressure altitude at which the fuel fow is a maximum.19. The maximum operating altitude for a certain aeroplane with a pressurized cabin:a. is dependent on aerodynamic ceiling.b. is dependent on the OAT.c. is only certifed for four-engine aeroplanes.d. is the highest pressure altitude certifed for normal operation.440Chapter 16Class AEn Route20. Why are step climbs used on long distance fights?a. Step climbs do not have any special purpose for jet aeroplanes; they are used for piston engine aeroplanes only.b. To respect ATC fight level constraints.c. To fy as close as possible to the optimum altitude as aeroplane mass reduces.d. Step climbs are only justifed if, at the higher altitude, less headwind or more tailwind can be expected.21. Long range cruise (LRC) instead of best range speed (MRC) is selected because LRC:a. is 4% faster and achieves 99% of maximum specifc range in zero wind.b. is the speed for best range.c. is the speed for best economy.d. gives higher specifc range with tailwind.22. What happens when an aeroplane climbs at a constant Mach number?a. IAS stays constant so there will be no problems.b. The 1.3G altitude is exceeded, so Mach bufet will start immediately.c. The lift coefcient increases.d. The TAS continues to increase, which may lead to structural problems.23. The speed range between low speed bufet and high speed bufet:a. decreases with increasing mass and is independent of altitude.b. is only limiting at low altitudes.c. increases with increasing mass.d. narrows with increasing mass and increasing altitude.24. A jet aeroplane descends with constant Mach number. Which of the following speed limits is most likely to be exceeded frst?a. Maximum Operational Mach Numberb. Maximum Operating Speedc. Never Exceed Speedd. High Speed Bufet Limit25. With respect to the optimum altitude, which of the following statements is correct?

a. An aeroplane always fies below the optimum altitude, because Mach bufet might occur.b. An aeroplane always fies at the optimum altitude because this is economically seen as the most atractive altitude.c. An aeroplane fies most of the time above the optimum altitude because this yields the most economic result.d. An aeroplane sometimes fies above or below the optimum altitude because optimum altitude increases continuously during fight.441Chapter 16Class AEn Route26. The aerodynamic ceiling:a. is the altitude at which the aeroplane reaches 50 ft/min.b. is the altitude at which the speeds for low speed bufet and for high speed bufet are the same.c. depends upon thrust seting and increase with increasing thrust.d. is the altitude at which the best rate of climb theoretically is zero.27. An aeroplane operating under the 180 minutes ETOPS rule may be up to:a. 180 minutes fying time to a suitable airport in still air with one engine inoperative.b. 180 minutes fying time to a suitable airport under the prevailing weather condition with one engine inoperative.c. 180 minutes fying time from suitable airport in still air at a normal cruising speed.d. 90 minutes fying time from the frst en-route airport and another 90 minutes from the second en-route airport in still air with one engine inoperative.28. If the climb speed schedule is changed from 280/.74 to 290/.74 the new crossover altitude is:a. unchanged.b. only afected by the aeroplane gross mass.c. lower.d. higher.29. The drift down procedure specifes requirements concerning the:a. engine power at the altitude at which engine failure occurs.b. climb gradient during the descent to the net level-of altitude.c. weight during landing at the alternate.d. obstacle clearance during descent to the net level-of altitude.30. For this question use Figure 4.24 in CAP 698 Section 4.With regard to the drift down performance of the twin jet aeroplane, why does the curve representing 35,000 kg gross mass in the chart for drift down net profles start at approximately 3 minutes at FL370?a. Because at this mass the engines slow down at a slower rate after failure, there is still some thrust left during four minutes.b. Due to higher TAS at this mass it takes more time to develop the optimal rate of descent, because of the inertia involved.c. All the curves start at the same point, which is situated outside the chart.d. Because at this mass it takes about 3 minutes to decelerate to the optimum speed for drift down at the original cruising level.31. Which of the following statements is correct?a. An engine failure at high cruising altitude will always result in a drift down, because it is not permited to fy the same altitude as with all engines operating.b. When determining the obstacle clearance during drift down, fuel dumping may be taken into account.c. The drift down regulations require a minimum descent angle after an engine failure at cruising altitude.d. The drift down procedure requires a minimum obstacle clearance of 35 ft.442Chapter 16Class AEn Route32. The drift down requirements are based on:a. the actual engine thrust output at the altitude of engine failure.b. the maximum fight path gradient during the descent.c. the landing mass limit at the alternate.d. the obstacle clearance during a descent to the new cruising altitude if an engine has failed.33. If the level-of altitude is below the obstacle clearance altitude during a drift down procedure?a. fuel jetisoning should be started at the beginning of drift down.b. the recommended drift down speed should be disregarded and it should be fown at the stall speed plus 10 kt.c. fuel jetisoning should be started when the obstacle clearance altitude is reached.d. the drift down should be fown with faps in the approach confguration.34. After engine failure the aeroplane is unable to maintain its cruising altitude. What is the procedure which should be applied?a. Emergency Descent Procedure.b. ETOPS.c. Long Range Cruise Descent.d. Drift Down Procedure.35. For this question use Figure 4.24 of CAP 698 Section 4.With regard to the drift down performance of the twin jet aeroplane, what is meant by equivalent gross weight at engine failure?a. The increment accounts for the higher fuel fow at higher temperatures.b. The equivalent gross weight at engine failure is the actual gross weight corrected for OAT higher than ISA +10C.c. The increment represents fuel used before engine failure.d. This gross weight accounts for the lower Mach number at higher temperatures.36. Drift down is the procedure to be applied:a. to conduct a visual approach if VASI is available.b. after engine failure if the aeroplane is above the one engine out maximum altitude.c. after cabin depressurization.d. to conduct an instrument approach at the alternate.37. In a twin engine jet aircraft with six passenger seats, and a maximum certifed take of mass of 5,650 kg. What is the required en-route obstacle clearance, with one engine inoperative during drift down towards the alternate airport?a. 2,000ftb. 1,500ftc. 1,000ftd. 50ft or half the wingspan443Chapter 16Class AEn Route38. Below the optimum cruise altitude:a. The IAS for long range cruise increases continuously with decreasing altitude.b. The Mach number for long range cruise decreases continuously with decreasing altitude.c. The Mach number for long range cruise decreases continuously with an increasing mass at a constant altitude.d. The Mach number for long range cruise increases continuously with decreasing mass at a constant altitude.39. How does the long range cruise speed (LRC) change?a. LRC Mach Number decreases with decreasing altitude.b. LRC Mach number decreases with increasing altitude.c. LRC Indicated airspeed increases with increasing altitude.d. LRC True airspeed decreases with increasing altitude.40 If a fight is performed with a higher cost index at a given mass, which of the following will occur?a. A beter maximum rangeb. A higher cruise mach numberc. A lower cruise mach numberd. A beter long range41. The efect of fying at the Long Range Cruise Speed instead of the Maximum Range Cruise Speed is:a. Fuel fow is reduced and speed stability is improved.b. Fuel fow is reduced and speed stability is reduced.c. Fuel fow is increased and speed stability is improved.d. Fuel fow is increased and speed stability is reduced.42. An aircrafts descent speed schedule is 0.74 M / 250 KIAS. During the descent from 30,000ft to sea level, the angle of atack will:a. Decrease, then remain constant.b. Increase, then remain constant.c. Remain constant.d. Decrease.43. During a drift down following engine failure, what would be the correct procedure to follow?a. Begin fuel jetison immediately, commensurate with having required reserves at destination.b. Do not commence fuel jetison until en-route obstacles have been cleared.c. Descend in the approach confguration.d. Disregard the fight manual and descend at Vs + 10kts to the destination.444Chapter 16Class AEn RouteANSWERS1 B 21 A 41 C2 C 22 C 42 A3 C 23 D 43 A4 A 24 B5 A 25 D6 B 26 B7 B 27 A8 D 28 C9 B 29 D10 A 30 D11 B 31 B12 A 32 D13 A 33 A14 B 34 D15 B 35 B16 B 36 B17 B 37 A18 C 38 B19 B 39 A20 C 40 B445Chapter 17Class ALandingCHAPTER SEVENTEENCLASS A: LANDINGContentsLANDING CONSIDERATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447LANDING CLIMB REQUIREMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . 447LANDING DISTANCE REQUIREMENTS JAR-OPS 1.515 . . . . . . . . . . . . . . . 449RUNWAY SELECTION / DESPATCH RULES . . . . . . . . . . . . . . . . . . . . . . 449PRESENTATION OF DATA. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 449QUESTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 451ANSWERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454446Chapter 17Class ALanding447Chapter 17Class ALandingLANDING CONSIDERATIONSThe maximum mass for landing is the lesser of: the climb limit mass the feld length limit mass the structural limit massLANDING CLIMB REQUIREMENTSLANDING CLIMB (All engines operating CS 25.119) A gradient of not less than 3.2% with: All engines operating at the power available 8 seconds after initiation of movement of the thrust control from the minimum fight idle to the take of position Landing confguration. Aerodrome altitude. Ambient temperature expected at the time of landing. A climb speed which is: not less than 1.13 VSR (may be 1.15 VSRR for 4 engine aircraft if application of power results in signifcant reduction in stalling speed) not less than VMCL not more than 1.23 VSRDISCONTINUED APPROACH CLIMB (One engine inoperative. CS 25.121) A climb gradient not less than: 2.1% for 2 engined aircraft 2.4% for 3 engined aircraft 2.7% for 4 engined aircraftWith: The critical engine inoperative and the remaining engines at the available take-ofthrust. Landing gear retracted Flaps approach confguration, provided that the approach fap Vs does not exceed 110% of landing fap Vs Aerodrome altitude Ambient temperature Speed: Normal approach speed but not greater than 1.41 VSR.The more limiting of the landing climb and the approach gradient requirements will determine the maximum mass for altitude and temperature at the landing aerodrome. Figure 17.1 shows a typical presentation of this data.DISCONTINUED APPROACH INSTRUMENT CLIMB. (JAR-OPS 1.510)For instrument approaches with decision heights below 200ft, an operator must verify that the approach mass of the aeroplane, taking into account the take-of mass and the fuel expected to be consumed in fight, allows a missed approach gradient of climb, with the critical engine failed and with the speed and confguration used for go-around of at least 2.5%, or the published gradient, whichever is the greater.448Chapter 17Class ALandingFigure 17.1. Landing Performance Climb Limit449Chapter 17Class ALandingLANDING DISTANCE REQUIREMENTS JAR-OPS 1.515The landing distance required on a dry runway for destination and alternate aerodromes, from 50 ft to a full stop must not exceed: 60% of the Landing Distance Available for turbo-jet aeroplanes 70% of the Landing Distance Available for turbo-prop aeroplanes(Short landing and steep approach procedures may be approved based on lower screen heights, but not less than 35 ft)The landing distance required is based on: the aeroplane in the landing confguration the speed at 50ft not less than 1.23 VSR aerodrome pressure altitude standard day temperature (ISA) factored winds (50% headwind, 150% tailwind) the runway slope if greater than 2%RUNWAY SELECTION / DESPATCH RULESLanding must be considered both in still air and in the forecast wind.a. Still air: The most favourable runway in still air may be selected. b. Forecast wind: The runway most likely to be used in the forecast wind. The lower of the two masses obtained from a. and b. above must be selected as the limiting mass for the feld lengths available.NON COMPLIANCE If the still air requirement cannot be met at an aerodrome with a single runway, that is, landing can only be made if there is an adequate wind component, the aircraft may be dispatched if 2 alternate aerodromes are designated at which full compliance is possible. If the forecast wind requirement cannot be met, the aeroplane may be dispatched if an alternate is designated at which all the landing requirements are met. WET RUNWAYSIf the runway is forecast to be wet at the estimated time of arrival the Landing Distance Available must be at least 115% of the dry runway landing distance required. However, a lesser factor may be used so long as it is published in the Aeroplane Flight Manual and the authority has approved such a factor.PRESENTATION OF DATAThe example graph in fgure 17.2 can be found in CAP 698 on page 46 of section 4. Work through the example shown in the graph which is detailed on page 40 of section 4. Practice by using the questions at the end of the chapter and remember that you will need to work though these graphs both in the normal way as illustrated by the arrow heads in the example, but also in reverse.450Chapter 17Class ALandingFigure 17.2 shows a typical presentation of landing distance data.451Chapter 17Class ALandingQUESTIONS1. The approach climb requirement has been established to ensure:a. minimum climb gradient in case of a go-around with one engine inoperative.b. obstacle clearance in the approach area.c. manoeuvrability in case of landing with one engine inoperative.d. manoeuvrability during approach with full faps and gear down, all engines operating.2. The maximum mass for landing could be limited by:a. the climb requirements with all engines in the landing confguration but with gear up.b. the climb requirements with one engine inoperative in the approach confguration.c. the climb requirements with one engine inoperative in the landing confguration.d. the climb requirements with all engines in the approach confguration.3. Which of the following is true according to JAA regulations for turbo propeller powered aeroplanes not performing a steep approach?a. Maximum Take-of Run is 0.5 x runway length.b. Maximum use of clearway is 1.5 x runway length.c. Maximum Landing Distance at the destination aerodrome and at any alternate aerodrome is 0.7 x LDA (Landing Distance Available).d. Maximum Landing Distance at destination is 0.95 x LDA (Landing Distance Available).4. For a turboprop powered aeroplane, a 2,200 m long runway at the destination aerodrome is expected to be wet. To ensure the wet landing distance meets the requirement, the dry runway landing distance should not exceed:a. 1,540 m.b. 1,147 m.c. 1,339 m.d. 1,771 m.5. A fight is planned with a turbojet aeroplane to an aerodrome with a landing distance available of 2,400 m. Which of the following is the maximum landing distance for a dry runway?a. 1 437 m.b. 1 250 m.c. 1 090 m.d. 1 655 m.6. The maximum mass for landing could be limited by:a. the climb requirements with all engines in the landing confguration but with gear up.b. the climb requirements with one engine inoperative in the approach confguration.c. the climb requirements with one engine inoperative in the landing confguration.d. the climb requirements with all engines in the approach confguration.452Chapter 17Class ALanding7. According to JAR-OPS 1, which one of the following statements concerning the landing distance for a turbojet aeroplane is correct?a. The landing distance is the distance from 35 ft above the surface of the runway to the full stop.b. When determining the maximum allowable landing mass at destination, 60% of the available landing runway length should be taken into account.c. Reverse thrust is one of the factors always taken into account when determining the landing distance required.d. Malfunctioning of an anti-skid system has no efect on the required runway length.8. By what factor must the landing distance available (dry runway) for a turbojet powered aeroplane be multiplied to fnd the maximum landing distance?a. 1.15b. 1.67c. 60/115d. 0.609. The landing feld length required for jet aeroplanes at the alternate in wet conditions is the demonstrated landing distance plus:a. 92%b. 43%c. 70%d. 67%10. The landing feld length required for turbojet aeroplanes at the destination in wet condition is the demonstrated landing distance plus:a. 67%b. 70%c. 43%d. 92%11. For a turbojet aeroplane, what is the maximum landing distance for wet runways when the landing distance available at an aerodrome is 3,000 m?a. 2 070 m.b. 1,562 m.c. 1,800 m.d. 2,609 m.12. If the airworthiness documents do not specify a correction for landing on a wet runway; the landing distance must be increased by:a. 10 %b. 20 %c. 15 %d. 5 %453Chapter 17Class ALanding13. Required runway length at destination airport for turboprop aeroplanes:a. is the same as at an alternate airport.b. is less then at an alternate airport.c. is more than at an alternate airport.d. is 60% longer than at an alternate airport.14. For this question use Figure 4.28 in CAP 698 Section 4.What is the minimum feld length required for the worst wind situation, landing a twin jet aeroplane with the anti-skid inoperative? Elevation: 2000 ft QNH: 1013 hPa Landing mass: 50,000 kg Flaps: as required for minimum landing distance Runway condition: dry Wind: Maximum allowable tailwind: 15 kt Maximum allowable headwind: 50 kta. 2600 m.b. 2700 m.c. 2900 m.d. 3100 m.15. Compared to the landing distance available, the maximum landing distance for a turbo-propeller and turbojet aircraft are:a. 60%, 60%b. 70%, 70%c. 70%, 60%d. 60%, 70%16. The landing climb gradient limit mass is determined by:a. A gradient of 3.2% with one engine inoperative, the other engines at take of power in the landing confguration.b. A gradient of 3.2% with all engines operating at take of power, in the landing confguration.c. A gradient of 2.1% with all engines operating at take of power, with landing gear retracted, and approach fap.d. A gradient of 3.2% with all engines operating at take of power, with landing gear retracted and approach fap.454Chapter 17Class ALandingANSWERS1. A2. B3. C4. C5. A6. B7. B8. D9. A10. D11. B12. C13. A14. D15. C16. B455Chapter 18 Revision QuestionsCHAPTER EIGHTEEN REVISION QUESTIONSContentsQUESTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .457ANSWERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .486SPECIMEN EXAMINATION PAPER . . . . . . . . . . . . . . . . . . . . . . . . . . . .487ANSWERS TO SPECIMEN EXAMINATION PAPER . . . . . . . . . . . . . . . . . . .494EXPLANTIONS TO ANSWERS SPECIMEN EXAM PAPER . . . . . . . . . . . . . .495456Revision Questions Chapter 18457Chapter 18 Revision QuestionsQUESTIONS1. What happens to the speed for Vx and Vy with increasing altitude?a. Both remain constant.b. Vx remains constant and Vy increases.c. Vx increases and Vy remains constant.d. Vx remains constant and Vy decreases.2. The afect of a contaminated runway on the feld limit mass :a. Decreased weight, increased V1, increased VR.b. Decreased weight, same V1, increased VR.c. Decreased weight, same V1, same VR.d. Decreased weight, decreased V1, decreased VR.3. When operating with anti-skid inoperative:a. Both landing and take of performance will be afected.b. Only landing performance will be afected.c. Only take of performance will be afected.d. Neither take of or landing performance will be afected.4. When comparing Vx to Vy:a. Vx will always be greater than Vy.b. Vy will always be greater than or equal to Vx.c. Vy will always be greater than Vx.d. Vx will sometimes be greater than Vy, but sometimes be less than Vy.5. Referring to Fig. 4.24. Why does the curve for an equivalent weight of 35 000 kg, only start 4 mins after engine failure?a. All the curves start at the same point higher up.b. At that altitude the engine takes longer to spool down after failure.c. At that weight the aircraft has a higher TAS and therefore more momentum.d. At that weight the aircraft takes longer to slowdown to the optimum drift down speed.6. With which conditions would one expect Vmc to be the lowest?a. Cold temp, low altitude, low humidity.b. Hot temp, low pressure altitude, high humidity.c. Hot temp, high pressure altitude, high humidity.d. Cold temp, high altitude, low humidity.7. Give the correct order for the following:a. Vmcg, VR, V1, V2b. Vmcg, V1, VR, V2c. V1, Vmcg, VR, V2d. Vmcg, V1, Vmca, VR, V2458Revision Questions Chapter 188. If the C of G moves aft from the most forward position:a. The range and the fuel consumption will increase.b. The range and the fuel consumption will decrease.c. The range will increase and the fuel consumption will decrease.d. The range will decrease and the fuel consumption will increase.9. Descending below the tropopause from FL370 to FL250 at a steady M#, then FL250 to FL100 at a constant CAS. What happens to descent angle?a. Increase Increaseb. Increase Constantc. Decrease Decreased. Constant Constant10. With a constant weight and M#, a higher altitude will require:a. Lower CL.b. Lower CD.c. Higher AoA.d. No change.11. When approaching a wet runway, with the risk of hydroplaning, what technique should the pilot adapt?a. Positive touch down, full reverse and brakes as soon as possible.b. Smoothest possible touch down, full reverse and only brakes below VP.c. Positive touch down, full reverse and only brakes below VP.d. Normal landing, full reverse and brakes at VP.12. An aircraft with a grad of 3.3%, flying at an IAS of 85 KTS. At a P.ALT of 8500 with a temp of +15c will have a ROC of:a. 284/Minb. 623/Minc. 1117/Mind. 334/Min12. An aircraft with a mass of 110000kg is capable of maintaining a grad of 2.6%. With all the atmospheric variables remaining the same, with what mass would it be able to achieve a grad of 2.4?a. 119167 kgb. 101530 kgc. 110000 kgd. 121167 kg14. Give the correct sequence:a. Vs, Vx, Vyb. Vx, Vs, Vyc. Vs, max range speed, max endurance speed.d. Max endurance speed, Vs, Max range speed.459Chapter 18 Revision Questions15. Flying at an altitude close to cofn corner gives:a. Max speed.b. Less manoeuvrability.c. Greater 1 engine inoperative Range.d. Greater 1 engine inoperative Endurance.16. The main reason for using the stepped climb technique is to:a. Decrease sector times.b. Increase endurance.c. Adhere to ATC procedures.d. Increase range.17. Ignoring the efect of compressibility, what would CL do with an increase in altitude?a. Increase.b. Decrease.c. Remain the same.d. Increase, then decrease.18. In climb limited mass calculations, the climb gradient is a ratio of:a. Height gained over distance travelled through the air.b. Height gained over distance travelled across the ground.c. TAS over rate of climb.d. TGS over rate of climb.19. In a twin engined jet aircraft with six passenger seats, and a maximum certifed take of mass of 5650kg. What is the required en-route obstacle clearance, with one engine inoperative during drift down towards the alternate airport?a. 2000ftb. 1500ftc. 1000ftd. 50ft or half the wingspan20. When does THRUST = DRAG?a. Climbing at a constant IAS.b. Descending at a constant IAS.c. Flying level at a constant IAS.d. All of the above.21. When take of mass is limited by VMBE, an increase in the uphill slope will:a. Have no afect.b. Require a decrease in the mass.c. Allow an increase in the mass.d. Decrease the TODR.460Revision Questions Chapter 1822. SFC will:a. Increase if C of G is moved further forward of the C of P.b. Decrease if C of G is moved further forward of the C of P.c. Not be afected by C of G position.d. Only be afected by C of G position, if it is behind the C of P.23. Reference point zero refers to:a. Point where the aircraft lifts of the ground.b. Point where the aircraft reaches V2.c. Point where the aircraft reaches 35ft.d. Point where gear is selected up.24. To maintain the same angle of atack and altitude at a higher gross weight an aeroplane needs:a. Less airspeed and same power.b. Same airspeed.c. More airspeed and less power.d. More airspeed and more power.25. The coefcient of lift may be increased by lowering the faps or:a. Increase CAS.b. Reduce nose up elevator trim.c. Increase angle of atack.d. Increase TAS.26. An aircraft is certifed to land with faps at either 25 or 35 degrees of fap. If the pilot selects the higher seting there will be:a. Increased landing distance and reduced go-around performance.b. Increased landing distance and improved go-around performance.c. Reduced landing distance and improved go-around performance.d. Reduced landing distance and reduced go-around performance.27. Which conditions are most suited to a selection of lower fap for take of?a. Low airfeld elevation, close obstacles, long runway, high temperature.b. Low airfeld elevation, no obstacles, short runway, low temperature.c. High elevation, no obstacles short runway, low temperature.d. High airfeld elevations, distant obstacles, long runway, high ambient temperature.28. During the certifcation of an aeroplane, the take of distance with all engines operating and the take of distance with one engine inoperative are 1547m 1720mWhat is the distance used in the aircraft certifcation?a. 1547mb. 1779mc. 1720md. 1798m461Chapter 18 Revision Questions29. V2 MIN is determined by: (excluding Vmca)a. 1.08VSR for 4 engine turboprops with 1.13VSR for 2 and 3 engined turboprops.b. 1.2VS for all turbojets.c. 1.2VSR for all turboprops and 1.15VSR for all turbojets.d. 1.15Vs for all aeroplanes.30. If the fap seting is changed from 10 degrees to 20 degrees V2 will:a. Not change.b. Decrease if not limited to Vmca.c. Increase.d. Increase or decrease depending on weight.31. For a turbojet aeroplane the second segment of the climb begins when:a. Accelerating from V2 to fap retraction speed begins.b. Landing gear is fully retracted.c. Flap retraction begins.d. Flaps are fully retracted.32. For a turbojet aeroplane the third segment of climb begins when:a. Acceleration to fap retraction speed begins (min 400ft).b. Landing gear is fully retracted.c. Acceleration from VLOF to V2 begins.d. Flaps are fully retracted.33. The bufet onset boundary chart tells the pilot:a. Critical mach number for various masses and altitudes.b. Values for low speed stall and mach bufet onset for various masses and altitudes.c. Mach number for low speed bufet and shock bufet for various masses and altitudes.d. Maximum operating MMO for various masses and altitudes.34. Two identical turbojets are at the same altitude and same speed and have the same specifc fuel consumption. Plane 1 weighs 130,000kg and fuel fow is 4,300kg/hr. If plane 2 weighs 115,000kg what is the fuel fow:a. 3804kg/hrb. 4044kg/hrc. 3364kg/hrd. 3530kg/hr35. The speed for minimum power required in a turbojet will be:a. Slower than the speed for minimum drag.b. Faster than the speed for minimum drag.c. Slower in a climb and faster in the decent.d. Same as speed for minimum drag.462Revision Questions Chapter 1836. In wet conditions, what extra percentage over the dry gross landing distance must be available for a turbojet?a. 43%b. 92%c. 67%d. 15%37. In dry conditions, when landing at an alternate airport in a turbojet by what factor should the landing distance available be divided by to give landing distance?a. 0.6b. 1.0c. 1.67d. 1.4338. What landing distance requirements need to be met at an alternate airfeld compared to a destination airfeld for a turboprop?a. Less than destination.b. More than destination.c. Same as destination.d. None applicable.39. For a twin engined aircraft, which can use either 5 or 15 degrees fap seting, using MRJT fg 4.4 what is the maximum feld limited take of mass: Pressure Altitude 7000 OAT -10C Length available 2400m Slope Level Wind Calma. 55000kgb. 56000kgc. 44000kgd. 52000kg40. Absolute Ceiling is defned by:a. Altitude where theoretical rate of climb is zero.b. Altitude at which rate of climb is 100fpm.c. Altitude obtained when using lowest steady fight speed.d. Altitude where low speed Bufet and high speed Bufet speeds are coincident.41. VREF for a Class B aircraft is defned by:a. 1.3Vsb. 1.2Vsc. 1.3 Vmcld. 1.2 Vmcl463Chapter 18 Revision Questions42. VR for a jet aircraft must be faster than, the greater of:a. 1.05 Vmca and V1b. Vmca and 1.1V1c. VMBE and V1d. V1 and 1.1 Vmca43. Landing on a runway with 5mm wet snow will:a. Increase landing distance.b. Decrease landing distance.c. Not afect the landing distance.d. Give a slightly reduced landing distance, due to increased impingement drag.44. Take of on a runway with standing water, with a depth of 0.5cm. Compared to a dry runway, feld length limited mass will:a. Increase, with a reduced V1.b. Remain the same, with a reduced V1.c. Decrease, with an increased V1.d. Decrease, with a decreased V1.45. A balanced feld length is when:a. Distance taken to accelerate to V1 and distance to stop are identicalb. TORA X 1.5c. V1 = VRd. ASDA equals TODA46. Increase ambient temperature will result in:a. Increased feld length limited mass.b. Decrease maximum brake energy limited mass.c. Increase climb limited mass.d. Increased obstacle limited mass.47. Pitch angle during decent at a constant mach number will:a. Increase.b. Decrease.c. Increase at frst then decrease.d. Stay constant.48. At maximum range speed in a turbojet the angle of atack is:a. Same as L/D max.b. Less than L/D max.c. Maximum.d. More than L/D max.464Revision Questions Chapter 1849. If there is an increase in atmospheric pressure and all other factors remain constant, it should result in:a. Decreased take of distance and increased climb performance.b. Increased take of distance and increased climb performance.c. Decreased take of distance and decreased climb performance.d. Increased take of distance and decreased climb performance.50. Climbing to cruise altitude with a headwind will:a. Increase time to climb.b. Decrease ground distance covered to climb.c. Decreased time to climb.d. Increased ground distance covered to climb.51. Requirements for the third segment of climb are:a. Minimum acceleration altitude for one engine inoperative should be used.b. There is no climb gradient requirement during acceleration phase.c. Level acceleration with an equivalent gradient of 1.2%.d. Legal minimum altitude for acceleration is 1500.52. If the calculations for an aeroplane of 3250lb indicate a service ceiling of 4000m, what will the service ceiling be when the actual take of mass is 3000lb?a. Higher.b. Lower.c. Higher or lower, more calculations will have to be done.d. The same.53. Why is there a requirement for an approach climb gradient?a. So that an aircraft falling below the glide path will be able to re-intercept it.b. Adequate performance for a go-around in the event of an engine failure.c. So that the aircraft will not stall when full fap is selected.d. To maintain minimum altitude on the approach.54. The drift down is a procedure applied:a. After aircraft depressurisation.b. For a visual approach to a VASI.c. For an instrument approach at an airfeld without an ILS.d. When the engine fails above the operating altitude for one engine inoperative.55. A light twin engined aircraft is climbing from the screen height of 50ft, and has an obstacle 10,000m along the net fight path. If the net climb gradient is 10%, there is no wind and obstacle is 900m above the aerodrome elevation then what will the clearance be?a. The aircraft will not clear the object.b. 85mc. 100md. 115m465Chapter 18 Revision Questions56. Take of run required for a jet aircraft, with one engine inoperative is:a. Brake release point to midpoint between VLOF and 35ft.b. Brake release point to 35ft.c. Brake release point to 15ft.d. The same as for all engines.57. A jet aircrafts maximum altitude is usually limited by:a. Its certifcation maximum altitude.b. Its pressurisation maximum altitude.c. The altitude at which low and high-speed bufet will occur.d. Thrust limits.58. With respect to en-route diversions (using drift down graph), if you believe that you will not clear an obstacle do you:a. Drift down to clearance height and then start to jetison fuel.b. Jetison fuel from the beginning of the drift down.c. Asses remaining fuel requirements, then jetison fuel as soon as possible.d. Fly slight faster.59. With respect to feld length limit, fll in the blanks in the follow statement. The distance to accelerate to ________, at which point an engine fails, followed by the reaction time of _________ and the ensuing deceleration to a full stop must be completed within the ____________.a. VR, 2sec, TORAb. V1, 2sec, ASDAc. VEF, 2sec, TORAd. VGO, 2sec, ASDA60. How does the power required graph move with an increase in altitude?a. Straight up.b. Straight down.c. Up and to the right.d. Straight across to the right.61. What factors would cause V2 to be limited by Vmca?a. Flaps at high setings.b. With high pressure.c. With low temperature.d. Combination of the above.62. In a climb, at a constant IAS / Mach No. 300 kts / 0.78 M. What happens at the change over point ( 29 500 ft, ISA ) ?a. Accelerate from the IAS to the Mach number, and therefore rate of climb will decrease.b. No change in rate of climb since TAS remains constant.c. Find that rate of climb would start to increase, because TAS starts to increase.d. Find that rate of climb would start to decrease, because TAS would start to decrease.466Revision Questions Chapter 1863. If not VMBE or Vmcg limited, what would V1 be limited by ?a. V2b. Vmgac. VRd. VMU64. What procedure is likely to require V1 to be reduced ?a. Improved climb procedure.b. Reduced thrust take of.c. When ASDA is greater than TODA.d. Take of with anti-skid inoperative.65. Which of the following is not afected by a tailwind :a. Landing climb limit mass.b. Obstacle limit mass.c. VMBE.d. Tyre speed limit mass.66. When fying an aircraft on the back of the drag curve, maintaining a slower speed (but still faster than VS) would require :a. More fap.b. Less thrust due to less parasite drag.c. More thrust.d. No change.67. During certifcation test fights for a turbojet aeroplane, the measured take of runs from brake release to a point equidistant between the point at which VLOF is reached and the point at which the aeroplane is 35ft above the take of surface are:1747 m with all engines operating.1950 m with the critical engine failure recognised at V1, other factors remaining unchanged.What is the correct value of the Take of Run?a. 1747 mb. 2243 mc. 1950 md. 2009 m68. Taking into account the following, what would be the minimum required head wind component for landing ? ( Using fg 2.4 in the CAP 698. )Factored landing distance of 1300 ft.ISA temperature at MSL.Landing mass of 3200 lbs.a. 10 Ktsb. 5 Ktsc. 0 Ktsd. 15 Kts467Chapter 18 Revision Questions69. Two identical aircraft at diferent masses are descending at idle thrust. Which of the following statements correctly describes their descent characteristics?a. There is no diference between the descent characteristics of the two aeroplanes.b. At a given angle of atack, the heavier aeroplane will always glide further than the lighter aeroplane.c. At a given angle of atack, the lighter aeroplane will always glide further than the heavier aeroplane.d. At a given angle of atack, both the vertical and the forward speeds are greater for the heavier aeroplane.70. When fying in a headwind, the speed for max range should be :a. slightly decreased.b. slightly increased.c. unchanged.d. should be increased, or decreased depending on the strength of the wind.71. VLO is defned as :a. Actual speed that the aircraft lifts of the ground.b. Minimum possible speed that the aircraft could lift of the ground.c. The maximum speed for landing gear operation.d. The long range cruise speed.72. When fying at the optimum range altitude, over time the :a. Fuel consumption gradually decreases.b. Fuel consumption gradually increases.c. Fuel consumption initially decreases then gradually increases.d. Fuel consumption remains constant.73. What happens to the feld limited take of mass with runway slope ?a. It increases with a downhill slope.b. It is unafected by runway slope.c. It decreases with a downhill slope.d. It increases with an uphill slope.74. For a given aircraft mass, fying with a cost index greater that zero set will result in :a. a cruise at a slower M# than the best range M# for a given altitude.b. a cruise at the maximum endurance speed.c. climb at the slowest safe speed, taking into account stall and speed stability.d. a cruise at a faster M# than the M# giving best ANM/kg ratio for a given altitude.75. Cruising with 1 or 2 engines inoperative at high altitude, compared to all engines operative cruise, range will :a. increase.b. decrease.c. not change.d. decrease with 1 engine inoperative, and increase with 2 engines inoperative.468Revision Questions Chapter 1876. Taking into account the values given below, what would be the maximum authorised brake release mass ? Flap : 5 10 15 Field limited mass : 49 850 kg 52 500 kg 56 850 kg Climb limited mass : 51 250 kg 49 300 kg 45 500 kga. 56 850 kgb. 49 300 kgc. 49 850 kgd. 51 250 kg77. A turbo-prop aircraft with a maximum all up mass in excess of 5700 kg, is limited to :a. 10 angle of bank up to 400 ft.b. 15 angle of bank up to 400 ft.c. 20 angle of bank up to 400 ft.d. 25 angle of bank up to 400 ft.78. With regards to the optimum altitude during the cruise, the aircraft is :a. always fown at the optimum altitude.b. always fown 2000 ft below the optimum altitude.c. may be fown above or below the optimum altitude, but never at the optimum altitude.d. fown as close to the optimum altitude as ATC will allow.79. A tailwind on take of will not afect :a. climb limit mass.b. obstacle clearance.c. feld limit mass.d. VMBE80. When climbing at a constant M# through the troposphere, TAS :a. increases.b. decreases.c. remains constant.d. increases then decreases.81. Concerning landing gear, which factors limit take of performance ?a. Brake temperature.b. Wheel rate of rotation, brake energy.c. Tyre temperature.d. Brake wear.82. In a glide ( power of descent ) if pitch angle is increased, glide distance will :a. increases.b. decrease.c. remain constant.d. depend on the aircraft.469Chapter 18 Revision Questions83. With which conditions would the aircraft need to be fown, in order to achieve maximum speed?a. Thrust set for minimum drag.b. Best lift - drag ratio.c. Maximum thrust and maximum drag.d. Maximum thrust and minimum drag.84. If a jet engine fails during take of, before V1 :a. the take of can be continued or aborted.b. the take of should be aborted.c. the take of should be continued.d. the take of may be continued if aircraft speed is, above VMCG and lies between VGO and VSTOP.85. Up to which height in NADP 1 noise abatement procedure must V2 + 10-20 Ktsbe maintained ?a. 1500 ft.b. 3000 ft.c. 1000 ft.d. 500 ft.86. At MSL, in ISA conditions. Climb gradient = 6 %What would the climb gradient be if : P.Altitude 1000 ft Temperature 17C Engine anti-ice on. Wing anti-ice on. ( - 0.2 % engine anti-ice, - 0,1 % wing anti ice, 0.2 % per 1000 ft P.altitude, 0.1 % per 1C ISA deviation )a. 5.1 %b. 6.3 %c. 3.8 %d. 5.5 %87. An aircraft with 180 minutes approval for ETOPS, must be :a. no more than 180 minutes from a suitable alternate, in still air, at the one engine inoperative TAS.b. no more than 180 minutes from a suitable alternate, in still air, at the all engine TAS.c. no more than 90 minutes from a suitable alternate, and 90 minutes from departure, at the one engine inoperative TASd. no more than 180 minutes from a suitable alternate, at the one engine inoperative TGS.88. In a balanced turn load factor is dependant on :a. radius of turn and aircraft weight.b. TAS and bank angle.c. radius of turn and bank angle.d. Bank angle only.470Revision Questions Chapter 1889. Puting in 16500 litres of fuel with an SG of 780 kg/m, and writing 16500 kg of fuel on the load sheet will result in :a. TOD increasing and ASD decreasing, and the calculated V2 being too fast.b. TOD and ASD decreasing, and the calculated V2 being too fast.c. TOD and ASD remaining constant, if the calculated speeds are used.d. TOD and ASD increasing, if the calculated speeds are used.90. If V1 is found to be lower than VMCG, which of the following statements will be true?a. VMCG must be reduced to equal V1.b. TOD will be greater than ASD.c. ASD will be greater than TOD.d. Take-of will not be permited.91. When gliding into a headwind airspeed should be :a. reduce airspeed to gust penetration speed.b. the same as the max. range glide speed in still air.c. lower than the max. range glide speed in still air.d. higher than the max. range glide speed in still air.92. How does the slush thickness afect the V1 reduction required ?a. Greater reduction if thicker.b. Smaller reduction if thicker.c. No afect if mass is reduced.d. No afect at all.93. Which denotes the stall speed in the landing confguration ?a. VSO.b. VS1.c. VS.d. VS1G.94. When in a gliding manoeuvre, in order to achieve maximum endurance the aircraft should be fown at :a. the speed for max. lift.b. the speed for min. drag.c. the speed for max. lift / drag.d. the speed for min. power.95. When descending below the optimum altitude at the long range cruise speed :a. Mach no. decreases.b. TAS increases.c. Mach no. remains constant.d. Mach no. increases.471Chapter 18 Revision Questions96. What does density altitude signify ?a. Pressure altitude.b. Flight levels.c. ISA altitude.d. An accurate indication of aircraft and engine performance.97. For a turbo-prop aircraft, the LDA at an aerodrome is 2200 m. If the conditions are indicated as wet, what would the equivalent dry LDA be ?a. 1451 m.b. 1913 m.c. 1538 m.d. 1317 m.98. During aircraft certifcation, the value of VMCG is found with nose wheel steering inoperative. This is because :a. Nose wheel steering does not afect VMCG.b. VMCG must be valid in both wet and dry conditions.c. Nose wheel steering does not work after an engine failure.d. The aircraft may be operated even if the nose wheel steering is inoperative.99. Referring to Fig 4.28.What would the landing distance required be for a MRJT aircraft with anti-skid inoperative if: Pressure altitude 2000 feet. Mass 50 000 kg Flaps for short feld. 15 Kts Tailwind Dry runway.a. 1700 mb. 2500 mc. 1900 md. 3100 m100. Which is true regarding a balanced feld ?a. Provides largest gap between net and gross margins.b. Provides minimum feld length required in the case of an engine failure.c. Take of distance will always be more than stopping distance.d. Distances will remain equal, even if engine failure speed is changed.101. Climbing in the troposphere at a constant TAS :a. Mach no. increases.b. Mach no. decreases.c. CAS increases.d. IAS increases.472Revision Questions Chapter 18102. When a MRJT aircraft descends at the maximum range speed :a. IAS increases.b. CAS increases.c. Mach no. decreases.d. Mach no. increases.103. What condition is found at the intersection of the thrust available and the drag curve :a. Un-accelerated fight in a climb.b. Accelerated climb.c. Un-accelerated level fight.d. Accelerated level fight.104. Out of the four forces acting on the aircraft in fight, what balances thrust in the climb ?a. Drag.b. Weightc. W Sin d. Drag + W Sin 105. According to the information in a light aircraft manual, which gives two power setings for cruise, 65 % and 75 %. If you fy at 75 % in stead of 65 % :a. Cruise speed will be higher, fuel consumption will be higher.b. Cruise speed will be the same, fuel consumption will be the same.c. Cruise speed will be higher, fuel consumption will be lower.d. Cruise speed will be higher, fuel consumption will be the same.106. With a downward sloping runway :a. V1 will increase.b. V1 will decrease.c. VR will increase.d. VR will decrease.107. How is fuel consumption afected by the C of G position, in terms of ANM / kg?a. Increases with a forward C of G.b. Decreases with a aft C of G.c. Decreases with a forward C of G.d. Fuel consumption is not afected by the C of G position.108. Rate of Climb 1000 ft/min TAS 198 ktsWhat is the aircrafts gradient ?a. 5.08 %b. 3 %c. 4 %d. 4.98 %473Chapter 18 Revision Questions109. The reduced thrust take of procedure may not be used when :a. Runway wet.b. After dark.c. Temperature varies by more than 10C from ISA.d. Anti-skid unserviceable.110. If the maximum take of mass is limited by tyre speed, what afect would a down sloping runway have ?a. No afect.b. Always increase the mass.c. Only increase the mass if not limited by any other limitation.d. Decrease the mass.111. With an obstacle which is 160 m above the airfeld elevation and 5000 m away from the end of the take of distance. ( Screen height 50 ft ) What would the obstacle clearance be with a gradient of 5% ?a. 105 mb. 90 mc. 250 md. 265 m112. Prior to take of the brake temperature needs to be checked, because :a. they indicate the state of the fusible plugs.b. if the brakes are already hot, they may fade / overheat during a RTO.c. they would work beter if they are warm.d. they may need to be warmed up to prevent them from cracking during a RTO.113. A turbo jet is fying at a constant M # in the cruise, how does SFC vary with OAT in Kelvin?a. Unrelated to Tb. Proportional to Tc. Proportional to 1/Td. Proportional to 1/T114. If an aircraft has a stall speed of 100 Kts, what would the speed on short fnals have to be ?a. 100 Ktsb. 115 Ktsc. 130 Ktsd. 120 Kts115. When descending at a constant M #, which speed is most likely to be exceeded frst ?a. Max operating speed.b. MMO.c. High speed bufet limit.d. VMO.474Revision Questions Chapter 18116. What is meant by Equivalent weight on the drift down profle graph ?a. Weight compensated for fuel reduction prior to engine failure.b. Weight compensated for temperature of ISA +10C and above.c. Weight compensated for density at diferent heights.d. Weight compensated for temperature at diferent heights.117. What happens to the speeds, VX and VY, when lowering the aircrafts undercarriage ?a. VX increases, VY decreases.b. VX decreases, VY decreases.c. VX increases, VY increases.d. VX decreases, VY increases.118. Maximum Endurance :a. can be achieved in level unaccelerated fight with minimum fuel consumption.b. can be achieved by fying at the best rate of climb speed in straight and level fight.c. can be achieved in a steady climb.d. can be achieved by fying at the absolute ceiling.119. What factors afect descent angle in a glide ?a. Confguration and altitude.b. Confguration and angle of atack.c. Mass and atitude.d. Mass and confguration.120. What is meant by balanced feld available ?a. TORA = TODAb. ASDA = ASDR and TODA =TODRc. TODA = ASDAd. TORA = ASDA121. For a piston-engined aeroplane at a constant altitude, angle ofatack and confguration, an increased weight will require :a. more power but less speed.b. more power and the same speed.c. more power and more speed.d. the same power but more speed.122. In the climb an aircraft has a thrust to weight ratio of 1:4 and a lift to drag ratio of 12: 1. While ignoring the slight diference between lift and weight in the climb, the climb gradient will be:a. 3.0 %b. 8.3 %c. 16.7 %d. 3.3 %475Chapter 18 Revision Questions123. Which of the following will not decrease the value of VS?a. The C of G in an aft position within the C of G envelope.b. Increased altitude.c. Decreased weight.d. Increased fap seting.124. All other factors being equal, the speed for minimum drag is:a. constant for all weights.b. a function of density altitude.c. proportional to weight.d. a function of pressure altitude.125. Taking into account the values given below, what would be the maximum authorised brake release mass with a 10 Kt tailwind?Flap : 5 10 15Field limited mass : 49 850 kg 52 500 kg 56 850 kgClimb limited mass : 51 250 kg 49 300 kg 45 500 kgAssume 370kg per Kt of tailwind.a. 56 850kgb. 49 850kgc. 52 500kgd. 48 800kg126. If a turn is commenced during the take of climb path :(i) The load factor(ii) The induced drag(iii) The climb gradient (i) (ii) (iii)a. Increases Decreases Decreasesb. Decreases Increases Increasesc. Increases Increases Decreasesd. Decreases Decreases Increases127. What efect does an increase in weight have on V1 ?a. It will cause it to increase.b. It will cause it to decrease.c. It will have no efect.d. It will cause it to decrease by the same percentage as the weight increase.128. VR for a Class A aeroplane must not be less than:a. 10 % above VMU.b. 5 % above VMCA.c. 5 % above VMCG.d. 10 % above VMCA.476Revision Questions Chapter 18129. As speed is reduced from VMD to VMP :a. power required decreases and drag decreases.b. power required decreases and drag increases.c. power required increases and drag increases.d. power required increases and drag decreases.130. The maximum induced drag occurs at a speed of :a. VMD.b. VMP.c. VSO.d. VATO.131. Profle drag is :a. inversely proportional to the square root of the EAS.b. directly proportional to the square of the EAS.c. inversely proportional to the square of the EAS.d. directly proportional to the square root of the EAS.132. VMD for a jet aeroplane is approximately equal to :a. 1.3 VS.b. 1.7 VS.c. 1.6 VS.d. 2.1 VS.133. The best EAS / Drag ratio is approximately :a. 1.3 VMD.b. 1.32 VMD.c. 1.6 VMD.d. 1.8 VMD.134. The efect an increase of weight has on the value of stalling speed (IAS) is that VS .a. increases.b. decreases.c. remains constant.d. increases or decreases, depending on the amount of weight increase.135. Which one of the following statements is true concerning the efect of changes of ambient temperature on an aeroplanes performance, assuming all other performance parameters remain constant ?a. An increase will cause a decrease in the landing distance required.b. An increase will cause a decrease in take of distance required.c. A decrease will cause an increase in the climb gradientd. A decrease will cause an increase in the take of ground run.477Chapter 18 Revision Questions136. What percentages of the head wind and tail wind components are taken into account, when calculating the take of feld length required ?a. 100% head wind and 100% tail wind.b. 150% head wind and 50% tail wind.c. 50% head wind and 100% tail wind.d. 50% head wind and 150% tail wind.137. For a turbo jet aircraft planning to land on a wet runway, the landing distance available :a. may be less than 15% greater than the dry landing distance if the fight manual gives specifc data for a wet runway.b. may be less than 15% greater than the dry landing distance if all reverse thrust systems are operative.c. may be less than 15% greater than the dry landing distance if permission is obtained from the relevant aerodrome authority.d. must always be at least 15% greater than the dry landing distance.138. The efect of installing more powerful engines in a turbojet aircraft is :a. to increase the aerodynamic ceiling and increase the performance ceiling.b. to decrease the aerodynamic ceiling and increase the performance ceiling.c. to increase the performance ceiling but not afect the aerodynamic ceiling.d. to decrease both the aerodynamic and the performance ceilings.139. In relation to runway strength, the ACN :a. may not exceed 90% of the PCN.b. may exceed the PCN by up to 10%.c. may never exceed the PCN.d. may exceed the PCN by a factor of 2.140. An aerodrome has a clearway of 500m and a stopway of 200m.If the stopway is extended to 500m the efect will be :a. the maximum take of mass will increase, and V1 will decrease.b. the maximum take of mass will increase and V1 will remain the same.c. the maximum take of mass will remain the same and V1 will increase.d. the maximum take of mass will increase and V1 will increase.141. An aircraft is climbing at a constant power seting and a speed of VX. If the speed is reduced and the power seting maintained the :a. Climb gradient will decrease and the rate of climb will increase.b. Climb gradient will decrease and the rate of climb will decrease.c. Climb gradient will increase and the rate of climb will increase.d. Climb gradient will increase and the rate of climb will decrease.142. An aircraft is climbing in a standard atmosphere above the tropopause at a constant Mach number :a. the IAS decreases and TAS remain constant.b. the IAS and TAS remain constant.c. the IAS decreases and TAS decreases.d. the IAS remains constant and TAS increases.478Revision Questions Chapter 18143. An aircraft is climbing at a constant IAS, below the Mach limit. As height increases :a. drag decreases, because density decreases.b. drag remains constant, but the climb gradient decreases.c. drag increases, because TAS increases.d. drag remains constant and the climb gradient remains constant.144. Optimum altitude can be defned as :a. the highest permissible altitude for an aeroplane type.b. the altitude at which an aeroplane atains the maximum specifc air range.c. the altitude at which the ground speed is greatest.d. the altitude at which specifc fuel consumption is highest.145. If an aircraft is descending at a constant Mach number :a. the IAS will increase and the margin to low speed bufet will decrease.b. the IAS will increase and the margin to low speed bufet will increase.c. the IAS will decrease and the margin to low speed bufet will decrease.d. the IAS will decrease and the margin to low speed bufet will increase.146. For a given fight level, the speed range determined by the bufet onset boundary chart will decrease with :a. decreased weight.b. decreased bank angle.c. a more forward CG position.d. increased ambient temperature.147. Which of the following variables will not afect the shape or position of the drag vs. IAS curve, for speeds below MCRIT ?a. Aspect ratio.b. Confguration.c. Altitude.d. Weight.148. Which of the following would give the greatest gliding endurance ?a. Flight at VMD.b. Flight at 1.32VMD.c. Flight at the best CL / CD ratio.d. Flight close to CL MAX.149. The tyre speed limit is :a. V1 in TAS.b. Max VLOF in TAS.c. Max VLOF in ground speed.d. V1 in ground speed.479Chapter 18 Revision Questions150. What gives one the greatest gliding time?a. Being light.b. A head wind.c. A tail wind.d. Being heavy.151. For take of performance calculations, what is taken into account ?a. OAT, pressure altitude, wind, weight.b. Standard temperature, altitude, wind, weight.c. Standard altitude, standard temperature, wind, weight.d. Standard temperature, pressure altitude, wind, weight.152. Which 3 speeds are efectively the same for a jet aircraft ?a. ROC, Range, minimum Drag.b. Range, Best angle of climb, minimum Drag.c. Best angle of climb, minimum Drag, Endurance.d. Best angle of climb, Range, Endurance.153. The long range cruise speed is a speed that gives :a. a 1% increase in range and a decrease in IAS.b. a 1% increase in TAS.c. a 1% increase in IAS.d. gives 99% of best cruise range, with an increase in IAS.154. When an aircraft takes-of at the mass it was limited to by the TODA:a. the end of the runway will be cleared by 35ft following an engine failure just before V1.b. the actual take-of mass equals the feld length limited take-of mass.c. the distance from BRP to V1 is equal to the distance from V1 to the 35ft screen.d. the balanced take-of distance equals 115% of the all engine take-of distance.155. Which of the following speeds give the maximum obstacle clearance in the climb?a. VYb. 1.2VS1c. VXd. VFE156. The tangent, from the origin to the Power required curve gives?a. Minimum Drag coefcient.b. L/D Minimumc. D/L Maximumd. L/D Maximum480Revision Questions Chapter 18157. For a jet fying at a constant altitude, at the maximum range speed, what is the efect on IAS and Drag over time?a. Increase, Increases.b. Decrease, Constant.c. Constant, Decrease.d. Decrease, Decrease.158. If an aircraft descends at a constant Mach #, what will the frst limiting speed be?a. Max operating speed.b. Never exceed speed.c. Max operating Mach #.d. Shock stall speed.159. For an aircraft gliding at its best glide range speed, if AoA is reduced:a. glide distance will increase.b. glide distance will remain unafected.c. glide distance will decrease.d. glide distance will remain constant, if speed is increased.160. If an aircrafts climb schedule was changed from 280 / 0.74 M to 290 / 0.74 M, what would happen to the change over altitude?a. It would remain unchanged.b. It could move up or down, depending on the aircraft.c. It will move down.d. It will move up.161. What happens to the cost index when fying above the optimum Long Range cruise speed?a. Cost index is not afected by speed.b. Cost index will increase with increased speed.c. Cost index will decrease with increased speed.d. It all depends on how much the speed is changed by.162. For an aircraft fying at the Long Range cruise speed, ( i ) Specifc Range and ( ii ) Fuel to time ratio:a) ( i ) Decreases ( ii ) Increasesb) ( i ) Increases ( ii ) Increasesc) ( i ) Decreases ( ii ) Decreasesd) ( i ) Increases ( ii ) Decreases163. By what percentage should V2 be greater than VMCA?a. 30%b. 10%c. 20%d. 15%481Chapter 18 Revision Questions164. If a turbo-prop aircraft has a wet LDA of 2200m, what would the equivalent dry landing distance allowed be?a. 1540mb. 1148mc. 1913md. 1339m165. If a TOD of 800m is calculated at sea level, on a level, dry runway, with standard conditions and with no wind, what would the TOD be for the conditions listed below? 2000 ft Airfeld elevation QNH 1013.25mb Temp. of 21C 5 kts of tailwind Dry runway with a 2% up slope.(Assuming: 20m/1000ft elevation, +10m/1kt of reported tailwind, 5m/1C ISA deviation andthe standard slope adjustments)a. 836mb. 940mc. 1034md. 1095m166. At a constant mass and altitude, a lower airspeed requires:a. more thrust and a lower coefcient of lift.b. less thrust and a lower coefcient of lift.c. more thrust and a lower coefcient of drag.d. a higher coefcient of lift.167. On a piston engine aeroplane, with increasing altitude at a constant gross mass, angle of atack and confguration, the power required:a. decreases slightly because of the lower air density.b. remains unchanged but the TAS increases.c. increases but the TAS remains constant.d. increases and the TAS increases.168. Reduced take of thrust:a. can be used if the headwind component during take of is at least 10 kts.b. can be used if the take of mass is higher than the performance limited take of mass.c. is not recommended at very low temperatures.d. has the beneft of improving engine life.169. Reduced take of thrust:a. can only be used in daylight.b. cant be used on a wet runway.c. is not recommended when wind shear is expected on departure.d. is not recommended at sea level.482Revision Questions Chapter 18170. May the anti-skid be considered in determining the take of and landing mass limits?a. Only landing.b. Only take of.c. Yesd. No171. Climb limited take-of mass can be increased by:a. Lower V2b. Lower fap seting and higher V2c. Lower VRd. Lower V1172. An operator shall ensure that the aircraft clears all obstacles in the net take-of fight path. The half-width of the Obstacle Accountability Area (Domain) at distance D from the end of the TODA is:a. 90m + (D / 0.125)b. 90m + (1.125 x D)c. 90m + (0.125 x D)d. (0.125 x D)173. The take-of performance for a turbo-jet aircraft using 10 Flap results in the following limitations:Obstacle clearance limited mass: 4 630 kgField length limited mass: 5 270 kg Given that it is intended to take-of with a mass of 5 000 kg, which of the following statements is true?a. With 5 fap the clearance limit will increase and the feld limit will decrease.b. With 15 fap both will increase.c. With 5 fap both will increase.d. With 15 fap the clearance limit will increase and the feld limit will decrease.174. Induced drag?a. Increases with increased airspeed.b. Decreases with increased airspeed.c. Independent of airspeed.d. Initially increases and the decreases with speed.483Chapter 18 Revision Questions175. Which of these graphs shows the relationship that thrust required has, with decreased weight?

487Chapter 18 Revision QuestionsSPECIMEN EXAMINATION PAPER40 Questions, 40 MarksTime Allowed: 1 hour1. A turbo-propeller aircraft is certifed with a maximum take-of mass of 5600 kg and a maximum passenger seating of 10. This aircraft would be certifed in:a. Class Ab. Class Bc. Class Cd. Either class A or class B depending on the number of passengers carried.(1 mark)2. How does the thrust from a fxed propeller change during the take-of run of an aircraft?a. It remains constant.b. It increases slightly as the aircraft speed builds up.c. It decreases slightly as the aircraft speed builds up.d. It only varies with changes in mass.(1 mark)3. The take-of run is defned as:a. distance to V1 and then to stop, assuing the engine failure is recognised at V1b. distance from brake release to the point where the aircraft reaches V2c. the horizontal distance from the start of the take-of roll to a point equidistant between VLOF and 35 ftd. the distance to 35 ft with an engine failure at V1 or 1.15 times the all engine distance to 35 ft.(1 mark)4. What efect does a downhill slope have on the take-of speeds?a. It has no efect on V1b. It decreases V1c. It increases V1d. It increases the IAS for take-of(1 mark)5. Which of the following combinations most reduces the take-of and climb performance of an aircraft?a. high temperature and high pressureb. low temperature and high pressurec. low temperature and low pressured. high temperature and low pressure(1 mark)488Revision Questions Chapter 186. Density altitude is:a. the true altitude of the aircraftb. the altitude in the standard atmosphere corresponding to the actual conditionsc. the indicated altitude on the altimeterd. used to calculate en-route safety altitudes(1 mark)7. The take-of climb gradient:a. increases in a head wind and decreases in a tail windb. decreases in a head wind and increases in a tail windc. is independent of the wind componentd. is determined with the aircraft in the take-of confguration(1 mark)8. The efect of changing altitude on the maximum rate of climb (ROC) and speed for best rate of climb for a turbo-jet aircraft, assuming everything else remains constant, is:a. as altitude increases the ROC and speed both decreaseb. as altitude increases the ROC and speed both increasec. as altitude increases the ROC decreases but the speed remains constantd. as altitude increases the ROC remains constant but the speed increases(1 mark)9. A runway at an aerodrome has a declared take-of run of 3000 m with 2000 m of clearway. The maximum distance that may be allowed for the take-of distance is:a. 5000mb. 6000 mc. 3000 md. 4500 m(1 mark)10. An aircraft may use either 5 or 15 fap seting for take of. The efect of selecting the 5 seting as compared to the 15 seting is:a. take-of distance and take-of climb gradient will both increaseb. take-of distance and take-of climb gradient will both decreasec. take-of distance will increase and take-of climb gradient will decreased. take-of distance will decrease and take-of climb gradient will increase(1 mark)11. The use of reduced thrust for take-of is permited:a. if the feld length limited take-of mass is greater than the climb limited take-of massb. if the actual take-of mass is less than the structural limiting massc. if the actual take-of mass is less than the feld length and climb limited take-ofmassesd. if the take-of distance required at the actual take-of mass does not exceed the take-of distance available(1 mark)489Chapter 18 Revision Questions12. Planning the performance for a runway with no obstacles, it is found that the climb limiting take-of mass is signifcantly greater than the feld limiting take-of mass with 5 fap selected. How can the limiting take-of mass be increased?a. use an increased V2 procedureb. increase the fap setingc. reduce the fap setingd. reduce the V2(1 mark)13. The maximum and minimum values of V1 are limited by:a. VR and VMCGb. V2 and VMCGc. VR and VMCAd. V2 and VMCA(1 mark)14. If the TAS is 175 kt and the rate of climb 1250 ft per minute, the climb gradient is approximately:a. 7%b. 14%c. 12%d. 10%(1 mark)15. A pilot inadvertently selects a V1 which is lower than the correct V1 for the actual take-of weight. What problem will the pilot encounter if an engine fails above the selected V1 but below the true V1?a. the accelerate-stop distance required will exceed the distance availableb. the climb gradient will be increasedc. the take-of distance required will exceed that availabled. there will be no signifcant efect on the performance(1 mark)16. A turbo-jet is in a climb at a constant IAS what happens to the drag?a. It increasesb. It decreasesc. it remains constantd. it increases initially then decreases(1 mark)17. Comparing the take-of performance of an aircraft from an aerodrome at 1000 ft to one taking of from an aerodrome at 6000 ft, the aircraft taking of from the aerodrome at 1000 ft:a. will require a greater take-of distance and have a greater climb gradientb. will require a greater take-of distance and have a lower climb gradientc. will require a shorter take-of distance and have a lower climb gradientd. will require a shorter take-of distance and have a greater climb gradient(1 mark)490Revision Questions Chapter 1818. Which is the correct sequence of speed?a. VS, VY, VXb. VX, VY, VXc. VS, VX, VYd. VX, VY, VS(1 mark)19. Which of the following will increase the accelerate-stop distance on a dry runway?a. a headwind componentb. an uphill slopec. temperatures below ISAd. low take-of mass, because of the increased acceleration(1 mark)20. A turbo-jet aircraft is climbing at a constant Mach number in the troposphere. Which of the following statements is correct?a. TAS and IAS increaseb. TAS and IAS decreasec. TAS decreases, IAS increasesd. TAS increases, IAS decreases(1 mark)21. The induced drag in an aeroplane:a. increases as speed increasesb. is independent of speedc. decreases as speed increasesd. decreases as weight decreases(1 mark)22. The speed range between low speed and high speed bufet:a. decreases as altitude increases and weight decreasesb. decreases as weight and altitude increasec. decreases as weight decreases and altitude increasesd. increases as weight decreases and altitude increases(1 mark)23. Thrust equals drag:a. in unaccelerated level fightb. in an unaccelerated descentc. in an unaccelerated climbd. in a climb, descent or level fight if unaccelerated(1 mark)491Chapter 18 Revision Questions24. A higher mass at a given altitude will reduce the gradient of climb and the rate of climb. But the speeds:a. VX and VY will decreaseb. VX and VY will increasec. VX will increase and VY will decreased. VX and VY will remain constant(1 mark)25. If the other factors are unchanged the fuel mileage (nm per kg) is:a. independent from the centre of gravity (CofG)b. lower with a forward CofGc. lower with an aft CofGd. higher with a forward CofG(1 mark)26. Concerning maximum range in a turbo-jet aircraft, which of the following is true?a. the speed to achieve maximum range is not afected by the wind componentb. to achieve maximum range speed should be increased in a headwind and reduced in a tailwindc. to achieve maximum range speed should be decreased in a headwind and increased in a tailwindd. The change is speed required to achieve maximum range is dependent on the strength of the wind component acting along the aircrafts fight path and may require either an increase or decrease for both headwind and tailwind(1 mark)27. V1 is the speed:a. above which take-of must be rejected if engine failure occursb. below which take-of must be continued if engine failure occursc. engine failure recognised below this speed, take-of must be rejected and above which take-of must be continuedd. the assumed speed for engine failure(1 mark)28. A constant headwind in the descent:a. Increases the angle of descentb. Increases the rate of descentc. Increases the angle of the descent fight pathd. Increases the ground distance travelled in the descent(1 mark)29. For a turbojet aircraft what is the reason for the use of maximum range speed?a. Greatest fight durationb. Minimum specifc fuel consumptionc. Minimum fight durationd. Minimum drag(1 mark)492Revision Questions Chapter 1830. Why are step climbs used on long range fights in jet transport aircraft?a. to comply with ATC fight level constraintsb. step climbs have no signifcance for jet aircraft, they are used by piston aircraftc. to fy as close as possible to the optimum altitude as mass reducesd. they are only justifed if the actual wind conditions difer signifcantly from the forecast conditions used for planning(1 mark)31. The absolute ceiling of an aircraft is:a. where the rate of climb reaches a specifed valueb. always lower than the aerodynamic ceilingc. where the rate of climb is theoretically zerod. where the gradient of climb is 5%(1 mark)32. In the take-of fight path, the net climb gradient when compared to the gross gradient is:a. greaterb. the samec. smallerd. dependent on aircraft type(1 mark)33. To answer this question use CAP698 SEP1 fgure 2.1. Conditions: aerodrome pressure altitude 1000 ft, temperature +30C, level, dry, concrete runway and 5 kt tailwind component. What is the regulated take-of distance to 50 ft for an aircraft weight 3500 lb if there is no stopway or clearway?a. 2800 ftb. 3220 ftc. 3640 ftd. 3500 ft(1 mark)34. To answer this question use CAP698 MRJT fgure 4.4. Conditions: Pressure altitude 5000 ft, temperature 5C, balanced feld length 2500 m, level runway, wind calm. What is the maximum feld length limited take-of mass and optimum fap seting?a. 59400 kg 15b. 60200 kg 5c. 59400 kg, 5d. 60200 kg 15(1 mark)35. The efect of a headwind component on glide range:a. the range will increaseb. the range will not be afectedc. the range will decreased. the range will only be afected if incorrect speeds are fown(1 mark)493Chapter 18 Revision Questions36. Refer to CAP398 MRJT fgure 4.24. At a mass of 35000 kg, why does the drift down curve start at approximately 3 minutes at an altitude of 37000 ft?a. The origin of the curve lies outside the chartb. At this altitude it takes longer for the engines to slow down, giving extra thrust for about 4 minutesc. Because of inertia at the higher TAS it takes longer to establish the optimum rate of descentd. It takes about this time to decelerate the aircraft to the optimum speed for drift down.(1 mark)37. A twin engine turbo-jet aircraft having lost one engine must clear obstacles in the drift down by a minimum of:a. 35 ftb. 1000 ftc. 1500 ftd. 2000 ft(1 mark)38. The landing speed, VREF, for a single engine aircraft must be not less than:a. 1.2VMCAb. 1.1VS0

c. 1.05VS0

d. 1.3VS0

(1 mark)39. What factor must be applied to the landing distance available at the destination aerodrome to determine the landing performance of a turbo-jet aircraft on a dry runway?a. 1.67b. 1.15c. 0.60d. 0.70(1 mark)40. An aircraft is certifed to use two landing fap positions, 25 and 35. If the pilot selects 25 instead of 35 then the aircraft will have:a. an increased landing distance and reduced go-around performanceb. an reduced landing distance and reduced go-around performancec. an increased landing distance and increased go-around performanced. an reduced landing distance and increased go-around performance(1 mark)494Revision Questions Chapter 18ANSWERS TO SPECIMEN EXAMINATION PAPER1 A 21 C2 C 22 B3 C 23 A4 B 24 B5 D 25 B6 B 26 B7 C 27 C8 A 28 C9 D 29 B10 A 30 C11 C 31 C12 B 32 C13 A 33 D14 A 34 D15 C 35 C16 C 36 D17 D 37 D18 C 38 D19 B 39 C20 B 40 C495Chapter 18 Revision QuestionsEXPLANTIONS TO ANSWERS SPECIMEN EXAM PAPERa. The requirement for an aircraft to be certifed in performance class B is that the maximum certifed take-of mass must not exceed 5700kg AND the certifed maximum number of passengers must not exceed 9. If both these conditions are not met then the aircraft will be certifed in class A.1. c. As speed increases then the angle of atack and hence the thrust of the propeller decrease.2. c.3. b. With a downhill slope the efort required to continue acceleration after VEF is less than the efort required to stop the aircraft because of the efect of gravity. Hence take-of can be achieved from a lower speed, but stopping the aircraft within the distance available can only be achieved from a lower speed.4. d. The lowest take-of performance will occur when air density is at its lowest, hence high temperature and low pressure.5. b. see defnitions6. c. In determining take-of climb performance, the still air gradient is considered.7. a. As altitude increases the speed for best gradient of climb remains constant, but the speed for best rate of climb decreases until the two speed coincide at the absolute ceiling.8. d. TODA is limited by the lower of TORA plus clearway and 1.5 x TORA.9. a. With a reduced fap seting the lift generated decreases and the stalling speed increases, so a greater take-of speed is required increasing the take-of distance. The reduced fap seting reduces the drag so the climb gradient increases.10. c. The use of reduced thrust means that the take-of distance will be increased and the climb gradient reduced, so neither of these can be limiting.11. b. Increasing the fap seting will reduce the TODR so the weight can be increased, but the climb gradient will reduce with the increased fap seting.12. a. see CAP698 page 62, paragraph 2.5.1.13. a. Gradient of climb can be approximated to rate of climb divided by TAS.14. c. The decision to continue the take-of will be made, but because the speed is below the normal V1, the distance required to accelerate will be greater, so the TODA may well be exceeded.15. c. At a constant IAS drag will remain constant.16. d. The air density is greater at the lower aerodrome so the performance will be beter in terms of acceleration and gradient.17. c. All speeds must be greater than Vs.496Revision Questions Chapter 1818. b. Although a uphill slope will enhance the deceleration, the efect on the acceleration to V1 will result in increased distances.19. b. Temperature is decreasing therefore the speed of sound and the TAS will decrease, and as altitude increases the IAS will also decrease.20. c. As speed increases the induced drag reduces proportional to the square of the speed increase.21. b. As altitude increases at a given weight the IAS for the onset of high speed bufet decreases and the IAS for the onset of low speed bufet remains constant, hence the speed range decreases. As weight increases at a given altitude the IAS for the onset of low speed bufet increases and there is a small decrease in the IAS for the onset of high speed bufet so once again the speed range decreases. 22. a. In the climb and descent an element of gravity acts along the aircraft axis either opposing (climb) or adding to (descent) the thrust, so only in level unaccelerated fight will the two forces be equal.23. b. As altitude increases the speed for best gradient of climb remains constant but the speed for the best rate of climb decreases. (see Q8)24. b. As the CofG moves forward the download on the tail increases, so the lift required also increases. To create the extra lift either speed or angle of atack must increase. In either case more thrust is required and the fuel required per nm will increase.25. b. The speed for maximum range in a jet aircraft is found where the line from the origin is tangential to the drag curve. With a head wind the origin moves to the right by the amount of the wind component so the line from the origin will be tangential at a higher speed, and to the left with a tail wind giving a lower speed.26. c. V1 is the decision speed. If engine failure is recognised below V1 take-of must be abandoned and engine failure recognised above V1 take-of must be continued.27. c. the rate of descent is independent of the wind, but the descent path is modifed by the wind, the angle increasing with a head wind because of the reduced ground distance covered.28. b. The maximum range speed is the speed at which the greatest distance can be fown, which means the lowest possible fuel usage. To achieve this implies we must have the lowest specifc fuel consumption.29. c. The most efcient way to operate a jet aircraft is to cruise climb, that is to set optimum cruise power and fy at the appropriate speed and allow the excess thrust as weight decreases to climb the aircraft.. For obvious safety reasons this is not possible, so the aircraft operates as close as possible to the optimum altitude by using the step climb technique.30. c. The absolute ceiling is the highest altitude to which the aircraft could be climbed, where, at optimum climb speed, thrust = drag, so with no excess of thrust the rate (and gradient) of climb will be zero.31. c. Net performance is gross (ie average) performance reduced by a regulatory amount to give a worst case view. Therefore the worst case for the climb gradient will be a lower gradient than it is expected to achieve.497Chapter 18 Revision Questions32. d. Gross take-of distance from the graph is 2800 ft. There is no stopway or clearway so the factor to apply is 1.25 to get the minimum TOR. (see CAP398, page 19)33. d. The maximum take-of mass will be achieved with the higher fap seting because the take-of speeds will be reduced. As ever take care when using the graphs.34. c. With a head wind component the glide performance will not be afected but the distance covered will reduce because of the reduced groundspeed.35. d. The aircraft will be allowed to slow down to the optimum drift down speed before commencing descent which will take an amount of time dependent on weight and altitude.36. d. This is the JAR regulatory requirement.37. d. Again the JAR regulatory requirement.38. c. The JAR regulatory requirement.39. c. With a reduced fap seting the stalling speed and hence the VREF will increase, so a greater landing distance will be required. With a reduced fap seting the drag will decrease, so the climb gradient will be beter.498Revision Questions Chapter 18